Giải SBT Toán 8 Cánh diều Bài 3. Phép nhân, phép chia phân thức đại số có đáp án
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Tính một cách hợp lí: a) 39x + 7/x - 2020.9x - 20/x + 2022 - 39x + 7/x - 2020.8x - 2042/x + 2022; b) x^2 - 81/x^2 + 101.( x^2 + 101/x - 9 + x^2 + 101/x + 9); c) x^2 - 1/x + 100.2x/x + 2 +

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Tính một cách hợp lí:

a) \(\frac{{39x + 7}}{{x - 2020}}.\frac{{9x - 20}}{{x + 2022}} - \frac{{39x + 7}}{{x - 2020}}.\frac{{8x - 2042}}{{x + 2022}}\);

b) \(\frac{{{x^2} - 81}}{{{x^2} + 101}}.\left( {\frac{{{x^2} + 101}}{{x - 9}} + \frac{{{x^2} + 101}}{{x + 9}}} \right)\);

c) \(\frac{{{x^2} - 1}}{{x + 100}}.\frac{{2x}}{{x + 2}} + \frac{{1 - {x^2}}}{{x + 100}}.\frac{{x - 100}}{{x + 2}}\).

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Lời giải

a) \(\frac{{39x + 7}}{{x - 2020}} \cdot \frac{{9x - 20}}{{x + 2022}} - \frac{{39x + 7}}{{x - 2020}} \cdot \frac{{8x - 2042}}{{x + 2022}}\)

\(\; = \frac{{39x + 7}}{{x - 2020}} \cdot \left( {\frac{{9x - 20}}{{x + 2022}} - \frac{{8x - 2042}}{{x + 2022}}} \right)\)

\(\; = \frac{{39x + 7}}{{x - 2020}} \cdot \frac{{9x - 20 - 8x + 2042}}{{x + 2022}}\)

\(\; = \frac{{39x + 7}}{{x - 2020}} \cdot \frac{{x + 2022}}{{x + 2022}} = \frac{{39x + 7}}{{x - 2020}}\).

b) \(\frac{{{x^2} - 81}}{{{x^2} + 101}}.\left( {\frac{{{x^2} + 101}}{{x - 9}} + \frac{{{x^2} + 101}}{{x + 9}}} \right)\)

\(\; = \frac{{\left( {x - 9} \right)\left( {x + 9} \right)}}{{{x^2} + 101}} \cdot \frac{{{x^2} + 101}}{{x - 9}} + \frac{{\left( {x - 9} \right)\left( {x + 9} \right)}}{{{x^2} + 101}} \cdot \frac{{{x^2} + 101}}{{x + 9}}\)

\( = \frac{{\left( {x - 9} \right)\left( {x + 9} \right)\left( {{x^2} + 101} \right)}}{{\left( {{x^2} + 101} \right)\left( {x - 9} \right)}} + \frac{{\left( {x - 9} \right)\left( {x + 9} \right)\left( {{x^2} + 101} \right)}}{{\left( {{x^2} + 101} \right)\left( {x + 9} \right)}}\)

= x + 9 + x ‒ 9 = 2x.

c) \(\frac{{{x^2} - 1}}{{x + 100}}.\frac{{2x}}{{x + 2}} + \frac{{1 - {x^2}}}{{x + 100}}.\frac{{x - 100}}{{x + 2}}\)

\( = \frac{{{x^2} - 1}}{{x + 100}}.\frac{{2x}}{{x + 2}} - \frac{{{x^2} - 1}}{{x + 100}}.\frac{{x - 100}}{{x + 2}}\)

\( = \frac{{{x^2} - 1}}{{x + 100}}.\left( {\frac{{2x}}{{x + 2}} - \frac{{x - 100}}{{x + 2}}} \right)\)

\( = \frac{{{x^2} - 1}}{{x + 100}}.\frac{{2x - x + 100}}{{x + 2}}\)

\( = \frac{{{x^2} - 1}}{{x + 100}}.\frac{{x + 100}}{{x + 2}} = \frac{{{x^2} - 1}}{{x + 2}}\).