Tính một cách hợp lí: a) 39x+7/x-2020 . 9x-20/x+2022 - 39x+7/x-2020 . 8x-20/x+2022

Giải SBT Toán 8 Bài 3: Phép nhân, phép chia phân thức đại số

Bài 16 trang 40 SBT Toán 8 Tập 1: 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:

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:

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