PAPER · REASONING · PRACTICE

Rational operations, integer powers and scientific notation

Consolidate common denominators, multiplication/division and negative powers, with short scientific-notation checks.

40 minutes · Year 8, upper volume · text pp. 83–96 · foundation lessons 42–48

Textbook reference: Year 8, upper · Chapter 15: rational expressions · lessons 42–48

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the upper-volume plan

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Teacher reference Arrange in My Classroom

Lesson routePhase

4′Check prerequisites
5′Observe and predict
8′Worked example and guided attempt
15′Independent practice
5′Discuss and correct
3′Exit questions

Observe · explain

A question before practising

Compare two correct routes for 1/x−1/(x+1). Why must the final restrictions survive either route?

Organise the key idea

Addition needs a common denominator; division needs a nonzero divisor. Negative exponents mean reciprocals, and scientific notation retains both sign and scale.

Open the original chapter

Use the lesson question here for paper discussion; the original activity opens separately.

Worked example1

Simplify and retain every restriction from the original expression.

\frac{1}{x-1}+\frac{2}{x+1}

Find values that make an original denominator or a divisor zero.
Use a common denominator and multiply the entire numerator by the required factor.
The simplified expression is (3x-1)/(x²-1); keep x≠-1, 1.

First record x≠-1, 1 from the original denominators and any divisor.
Use a common denominator and multiply the entire numerator by the required factor.
(1)/(x-1)+(2)/(x+1) = ((x+1)+2(x-1))/((x-1)(x+1)).
Therefore (3x-1)/(x²-1), with x≠-1, 1 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Guided attempt2

Simplify and retain every restriction from the original expression.

\frac{x^2-4}{x+1}\div\frac{x-2}{x+1}

Original denominators exclude ___; the divisor also excludes ___.
Replace÷ by× the reciprocal, factor, then cancel ___.

Find values that make an original denominator or a divisor zero.
Replace division by multiplication by the reciprocal; the original divisor must remain nonzero.
The simplified expression is x+2; keep x≠-1, 2.

First record x≠-1, 2 from the original denominators and any divisor.
Replace division by multiplication by the reciprocal; the original divisor must remain nonzero.
(x²-4)/(x+1)÷(x-2)/(x+1) = ((x-2)(x+2))/(x+1)×(x+1)/(x-2).
Therefore x+2, with x≠-1, 2 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Prerequisite3

Evaluate 2⁻³. Explain the negative exponent.

2^{-3}

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

A negative exponent means a reciprocal, not a negative value.
2⁻³=1/2³=1/8.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Prerequisite4

Simplify and retain every restriction from the original expression.

\frac{x+1}{x-2}+\frac{2}{x-2}

Find values that make an original denominator or a divisor zero.
The denominators already match; add the numerators only.
The simplified expression is (x+3)/(x-2); keep x≠2.

First record x≠2 from the original denominators and any divisor.
The denominators already match; add the numerators only.
(x+1)/(x-2)+(2)/(x-2) = (x+1+2)/(x-2).
Therefore (x+3)/(x-2), with x≠2 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Independent practice5

Simplify and retain every restriction from the original expression.

\frac{x^2-9}{x-1}\times\frac{x-1}{x+3}

Find values that make an original denominator or a divisor zero.
Factor first, then cancel common factors; never cancel separate added terms.
The simplified expression is x-3; keep x≠1, -3.

First record x≠1, -3 from the original denominators and any divisor.
Factor first, then cancel common factors; never cancel separate added terms.
(x²-9)/(x-1)×(x-1)/(x+3) = ((x-3)(x+3)(x-1))/((x-1)(x+3)).
Therefore x-3, with x≠1, -3 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Independent practice6

Simplify and retain every restriction from the original expression.

\frac{2}{x}-\frac{1}{x+1}

Find values that make an original denominator or a divisor zero.
Use a common denominator and multiply the entire numerator by the required factor.
The simplified expression is (x+2)/(x(x+1)); keep x≠0, -1.

First record x≠0, -1 from the original denominators and any divisor.
Use a common denominator and multiply the entire numerator by the required factor.
(2)/(x)-(1)/(x+1) = (2(x+1)-x)/(x(x+1)).
Therefore (x+2)/(x(x+1)), with x≠0, -1 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Independent practice7

Evaluate (−2)⁻². Show the reciprocal before calculating.

(-2)^{-2}

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

The whole negative base is squared after taking the reciprocal.
(−2)⁻²=1/(−2)²=1/4.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice8

Write 0.00056 in scientific notation with 1≤|a|<10.

0.00056

Keep any negative sign; it does not change the decimal shift count.
For a number with absolute value below 1, the power of 10 is negative.
Use 5.6×10⁻⁴; check by converting back.

Move the decimal point to obtain mantissa 5.6, whose absolute value is between 1 and 10.
Compensate with 10⁻⁴, so 0.00056=5.6×10⁻⁴.
Check the sign and the decimal order of magnitude.

Check your work

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Exit question9

Simplify and retain every restriction from the original expression.

\frac{3}{x}\div\frac{6}{x}

Find values that make an original denominator or a divisor zero.
Replace division by multiplication by the reciprocal; the original divisor must remain nonzero.
The simplified expression is (1)/(2); keep x≠0.

First record x≠0 from the original denominators and any divisor.
Replace division by multiplication by the reciprocal; the original divisor must remain nonzero.
(3)/(x)÷(6)/(x) = (3)/(x)×(x)/(6).
Therefore (1)/(2), with x≠0 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Exit question10

Write 0.000072 in scientific notation with 1≤|a|<10.

0.000072

Keep any negative sign; it does not change the decimal shift count.
For a number with absolute value below 1, the power of 10 is negative.
Use 7.2×10⁻⁵; check by converting back.

Move the decimal point to obtain mantissa 7.2, whose absolute value is between 1 and 10.
Compensate with 10⁻⁵, so 0.000072=7.2×10⁻⁵.
Check the sign and the decimal order of magnitude.

Check your work

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Further practice11

Simplify and retain every restriction from the original expression.

(\frac{1}{x}-\frac{1}{x+1})(x^2+x)

Find values that make an original denominator or a divisor zero.
Complete the subtraction first, factor x²+x=x(x+1), then multiply and cancel.
The simplified expression is 1; keep x≠0, -1.

First record x≠0, -1 from the original denominators and any divisor.
Complete the subtraction first, factor x²+x=x(x+1), then multiply and cancel.
((1)/(x)-(1)/(x+1))(x²+x) = (1)/(x(x+1))x(x+1).
Therefore 1, with x≠0, -1 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Further practice12

Simplify and retain every restriction from the original expression.

\frac{x+2}{x}-\frac{2}{x}

Find values that make an original denominator or a divisor zero.
Subtract the entire second numerator: (x+2)−2=x.
The simplified expression is 1; keep x≠0.

First record x≠0 from the original denominators and any divisor.
Subtract the entire second numerator: (x+2)−2=x.
(x+2)/(x)-(2)/(x) = (x+2-2)/(x).
Therefore 1, with x≠0 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Further practice13

Write -0.00032 in scientific notation with 1≤|a|<10.

-0.00032

Keep any negative sign; it does not change the decimal shift count.
For a number with absolute value below 1, the power of 10 is negative.
Use -3.2×10⁻⁴; check by converting back.

Move the decimal point to obtain mantissa -3.2, whose absolute value is between 1 and 10.
Compensate with 10⁻⁴, so -0.00032=-3.2×10⁻⁴.
Check the sign and the decimal order of magnitude.

Check your work

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Further practice14

Evaluate (−2/3)⁻².

(-2/3)^{-2}

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

Invert the base and change the exponent to positive 2.
(−3/2)²=9/4.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Extension15

Simplify and retain every restriction from the original expression.

(\frac{1}{x-1}-\frac{1}{x+1})\div\frac{2}{x^2-1}

Find values that make an original denominator or a divisor zero.
The bracket simplifies to 2/(x²−1), then divides by that same nonzero expression.
The simplified expression is 1; keep x≠-1, 1.

First record x≠-1, 1 from the original denominators and any divisor.
The bracket simplifies to 2/(x²−1), then divides by that same nonzero expression.
((1)/(x-1)-(1)/(x+1))÷(2)/(x²-1) = (2)/(x²-1)÷(2)/(x²-1).
Therefore 1, with x≠-1, 1 still required.

Check your work

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Extension16

Calculate 0.0003×0.0002 and express the product in scientific notation.

0.0003\times0.0002

Keep any negative sign; it does not change the decimal shift count.
For a number with absolute value below 1, the power of 10 is negative.
Use 6×10⁻⁸; check by converting back.

0.0003=3×10⁻⁴ and 0.0002=2×10⁻⁴.
Multiply the mantissas and add the exponents:3×2×10⁻⁸=6×10⁻⁸.
The mantissa 6 is normalized; the product is 0.00000006.

Check your work

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Preparation and assessment notes

Prerequisites

Factorisation, domain restrictions, fraction arithmetic and positive power laws.

Look for in students’ work

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Print edition

Textbook reference and authorship

Chapter-core practice after the mapped textbook lessons; these tasks do not replace every first-teaching period.

大湾区初中数学三维课堂（八年级上册）正文.pdf

Year 8, upper volume · text pp. 83–96 · foundation lessons 42–48

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Rational operations, integer powers and scientific notation

Year 8, upper volume · text pp. 83–96 · foundation lessons 42–48 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

Simplify and retain every restriction from the original expression.

\frac{1}{x-1}+\frac{2}{x+1}

Guided attempt2

Simplify and retain every restriction from the original expression.

\frac{x^2-4}{x+1}\div\frac{x-2}{x+1}

Original denominators exclude ___; the divisor also excludes ___.
Replace÷ by× the reciprocal, factor, then cancel ___.

Prerequisite3

Evaluate 2⁻³. Explain the negative exponent.

2^{-3}

Prerequisite4

Simplify and retain every restriction from the original expression.

\frac{x+1}{x-2}+\frac{2}{x-2}

Rational operations, integer powers and scientific notation

Year 8, upper volume · text pp. 83–96 · foundation lessons 42–48 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

Simplify and retain every restriction from the original expression.

\frac{x^2-9}{x-1}\times\frac{x-1}{x+3}

Independent practice6

Simplify and retain every restriction from the original expression.

\frac{2}{x}-\frac{1}{x+1}

Independent practice7

Evaluate (−2)⁻². Show the reciprocal before calculating.

(-2)^{-2}

Independent practice8

Write 0.00056 in scientific notation with 1≤|a|<10.

0.00056

Exit question9

Simplify and retain every restriction from the original expression.

\frac{3}{x}\div\frac{6}{x}

Exit question10

Write 0.000072 in scientific notation with 1≤|a|<10.

0.000072

Rational operations, integer powers and scientific notation — Further practice

Year 8, upper volume · text pp. 83–96 · foundation lessons 42–48

Further practice11

Simplify and retain every restriction from the original expression.

(\frac{1}{x}-\frac{1}{x+1})(x^2+x)

Further practice12

Simplify and retain every restriction from the original expression.

\frac{x+2}{x}-\frac{2}{x}

Further practice13

Write -0.00032 in scientific notation with 1≤|a|<10.

-0.00032

Further practice14

Evaluate (−2/3)⁻².

(-2/3)^{-2}

Extension15

Simplify and retain every restriction from the original expression.

(\frac{1}{x-1}-\frac{1}{x+1})\div\frac{2}{x^2-1}

Extension16

Calculate 0.0003×0.0002 and express the product in scientific notation.

0.0003\times0.0002

Rational operations, integer powers and scientific notation — Teacher answers

Year 8, upper volume · text pp. 83–96 · foundation lessons 42–48

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

Worked example1

Simplify and retain every restriction from the original expression.

\frac{1}{x-1}+\frac{2}{x+1}

Reference answer

(3x-1)/(x²-1); x≠-1, 1.

\frac{3x-1}{x^2-1}

First record x≠-1, 1 from the original denominators and any divisor.
Use a common denominator and multiply the entire numerator by the required factor.
(1)/(x-1)+(2)/(x+1) = ((x+1)+2(x-1))/((x-1)(x+1)).
Therefore (3x-1)/(x²-1), with x≠-1, 1 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Guided attempt2

Simplify and retain every restriction from the original expression.

\frac{x^2-4}{x+1}\div\frac{x-2}{x+1}

Original denominators exclude ___; the divisor also excludes ___.
Replace÷ by× the reciprocal, factor, then cancel ___.

Reference answer

x+2; x≠-1, 2.

x+2

First record x≠-1, 2 from the original denominators and any divisor.
Replace division by multiplication by the reciprocal; the original divisor must remain nonzero.
(x²-4)/(x+1)÷(x-2)/(x+1) = ((x-2)(x+2))/(x+1)×(x+1)/(x-2).
Therefore x+2, with x≠-1, 2 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Prerequisite3

Evaluate 2⁻³. Explain the negative exponent.

2^{-3}

Reference answer

1/8.

\frac{1}{8}

A negative exponent means a reciprocal, not a negative value.
2⁻³=1/2³=1/8.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Prerequisite4

Simplify and retain every restriction from the original expression.

\frac{x+1}{x-2}+\frac{2}{x-2}

Reference answer

(x+3)/(x-2); x≠2.

\frac{x+3}{x-2}

First record x≠2 from the original denominators and any divisor.
The denominators already match; add the numerators only.
(x+1)/(x-2)+(2)/(x-2) = (x+1+2)/(x-2).
Therefore (x+3)/(x-2), with x≠2 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Independent practice5

Simplify and retain every restriction from the original expression.

\frac{x^2-9}{x-1}\times\frac{x-1}{x+3}

Reference answer

x-3; x≠1, -3.

x-3

First record x≠1, -3 from the original denominators and any divisor.
Factor first, then cancel common factors; never cancel separate added terms.
(x²-9)/(x-1)×(x-1)/(x+3) = ((x-3)(x+3)(x-1))/((x-1)(x+3)).
Therefore x-3, with x≠1, -3 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Independent practice6

Simplify and retain every restriction from the original expression.

\frac{2}{x}-\frac{1}{x+1}

Reference answer

(x+2)/(x(x+1)); x≠0, -1.

\frac{x+2}{x(x+1)}

First record x≠0, -1 from the original denominators and any divisor.
Use a common denominator and multiply the entire numerator by the required factor.
(2)/(x)-(1)/(x+1) = (2(x+1)-x)/(x(x+1)).
Therefore (x+2)/(x(x+1)), with x≠0, -1 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Independent practice7

Evaluate (−2)⁻². Show the reciprocal before calculating.

(-2)^{-2}

Reference answer

1/4.

\frac{1}{4}

The whole negative base is squared after taking the reciprocal.
(−2)⁻²=1/(−2)²=1/4.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice8

Write 0.00056 in scientific notation with 1≤|a|<10.

0.00056

Reference answer

5.6×10⁻⁴.

5.6\times10^{-4}

Move the decimal point to obtain mantissa 5.6, whose absolute value is between 1 and 10.
Compensate with 10⁻⁴, so 0.00056=5.6×10⁻⁴.
Check the sign and the decimal order of magnitude.

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Exit question9

Simplify and retain every restriction from the original expression.

\frac{3}{x}\div\frac{6}{x}

Reference answer

(1)/(2); x≠0.

\frac{1}{2}

First record x≠0 from the original denominators and any divisor.
Replace division by multiplication by the reciprocal; the original divisor must remain nonzero.
(3)/(x)÷(6)/(x) = (3)/(x)×(x)/(6).
Therefore (1)/(2), with x≠0 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Exit question10

Write 0.000072 in scientific notation with 1≤|a|<10.

0.000072

Reference answer

7.2×10⁻⁵.

7.2\times10^{-5}

Move the decimal point to obtain mantissa 7.2, whose absolute value is between 1 and 10.
Compensate with 10⁻⁵, so 0.000072=7.2×10⁻⁵.
Check the sign and the decimal order of magnitude.

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Further practice11

Simplify and retain every restriction from the original expression.

(\frac{1}{x}-\frac{1}{x+1})(x^2+x)

Reference answer

1; x≠0, -1.

1

First record x≠0, -1 from the original denominators and any divisor.
Complete the subtraction first, factor x²+x=x(x+1), then multiply and cancel.
((1)/(x)-(1)/(x+1))(x²+x) = (1)/(x(x+1))x(x+1).
Therefore 1, with x≠0, -1 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Further practice12

Simplify and retain every restriction from the original expression.

\frac{x+2}{x}-\frac{2}{x}

Reference answer

1; x≠0.

1

First record x≠0 from the original denominators and any divisor.
Subtract the entire second numerator: (x+2)−2=x.
(x+2)/(x)-(2)/(x) = (x+2-2)/(x).
Therefore 1, with x≠0 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Further practice13

Write -0.00032 in scientific notation with 1≤|a|<10.

-0.00032

Reference answer

-3.2×10⁻⁴.

-3.2\times10^{-4}

Move the decimal point to obtain mantissa -3.2, whose absolute value is between 1 and 10.
Compensate with 10⁻⁴, so -0.00032=-3.2×10⁻⁴.
Check the sign and the decimal order of magnitude.

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.

Further practice14

Evaluate (−2/3)⁻².

(-2/3)^{-2}

Reference answer

9/4.

\frac{9}{4}

Invert the base and change the exponent to positive 2.
(−3/2)²=9/4.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Extension15

Simplify and retain every restriction from the original expression.

(\frac{1}{x-1}-\frac{1}{x+1})\div\frac{2}{x^2-1}

Reference answer

1; x≠-1, 1.

1

First record x≠-1, 1 from the original denominators and any divisor.
The bracket simplifies to 2/(x²−1), then divides by that same nonzero expression.
((1)/(x-1)-(1)/(x+1))÷(2)/(x²-1) = (2)/(x²-1)÷(2)/(x²-1).
Therefore 1, with x≠-1, 1 still required.

Use factor cancellation or a correct common denominator.
Retain all original restrictions, including a nonzero divisor.

Extension16

Calculate 0.0003×0.0002 and express the product in scientific notation.

0.0003\times0.0002

Reference answer

6×10⁻⁸.

6\times10^{-8}

0.0003=3×10⁻⁴ and 0.0002=2×10⁻⁴.
Multiply the mantissas and add the exponents:3×2×10⁻⁸=6×10⁻⁸.
The mantissa 6 is normalized; the product is 0.00000006.

Use a normalized nonzero mantissa and an integer exponent.
Preserve the original value and sign.