Giải bài 16 trang 40 sách bài tập toán 8 - Cánh diều

Đề bài

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

Lời giải chi tiết

a) Ta có:

\(\begin{array}{l}\frac{{39x + 7}}{{x - 2020}}.\frac{{9x - 20}}{{x + 2022}} - \frac{{39x + 7}}{{x - 2020}}.\frac{{8x - 2042}}{{x + 2022}}\\ = \frac{{39 + 7}}{{x - 2020}}.\left( {\frac{{9x - 20}}{{x + 2022}} - \frac{{8x - 2042}}{{x + 2022}}} \right)\\ = \frac{{39 + 7}}{{x - 2020}}.\frac{{x + 2022}}{{x + 2022}}\\ = \frac{{39 + 7}}{{x - 2020}}\end{array}\)

b) Ta có:

\(\begin{array}{l}\frac{{{x^2} - 81}}{{{x^2} + 101}}.\left( {\frac{{{x^2} + 101}}{{x - 9}} + \frac{{{x^2} + 101}}{{x + 9}}} \right)\\ = \frac{{\left( {x - 9} \right)\left( {x + 9} \right)}}{{{x^2} + 101}}.\frac{{{x^2} + 101}}{{x - 9}} + \frac{{\left( {x - 9} \right)\left( {x + 9} \right)}}{{{x^2} + 101}}.\frac{{{x^2} + 101}}{{x + 9}}\\ = x + 9 + x - 9 = 2x\end{array}\)

c) Ta có:

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