Rút gọn các biểu thức sau: [(9/x^3-9x)+(1/x+23)]/[(x-3/x^2+3x)-(x/3x+9)

Giải SBT Toán 8 (Kết nối tri thức) Bài 24: Phép nhân và phép chia phân thức đại số

Bài tập 6.31 trang 13 SBT Toán 8 Tập 2: Rút gọn các biểu thức sau

a) $\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)$;

b) $\frac{x+1}{x+2}.\left(\frac{x+2}{x+3}:\frac{x+3}{x+1}\right)$.

Lời giải:

a) $\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)$

$=\left.\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right]:\left.\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]$

$=\left.\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{1}{x+3}\right]:\left.\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right]$

$=\left.\frac{9}{x\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right]:\left.\frac{3\left(x-3\right)}{3x\left(x+3\right)}-\frac{x^2}{3x\left(x+3\right)}\right]$

$=\left.\frac{9+x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}\right]:\left.\frac{3\left(x-3\right)-x^2}{3x\left(x+3\right)}\right]$

$=\frac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\frac{3x-9-x^2}{3x\left(x+3\right)}$

$=\frac{x^2-3x+9}{x\left(x-3\right)\left(x+3\right)}.\frac{3x\left(x+3\right)}{-\left(x^2-3x+9\right)}$

$=\frac{-3}{x-3}$

b) $\frac{x+1}{x+2}.\left(\frac{x+2}{x+3}:\frac{x+3}{x+1}\right)$

$=\frac{x+1}{x+2}.\left(\frac{x+2}{x+3}.\frac{x+1}{x+3}\right)$

$=\frac{x+1}{x+2}.\frac{\left(x+2\right).\left(x+1\right)}{{\left(x+3\right)}^2}$

$=\frac{\left(x+1\right)\left(x+2\right).\left(x+1\right)}{\left(x+2\right){\left(x+3\right)}^2}$

$=\frac{{\left(x+1\right)}^2}{{\left(x+3\right)}^2}={\left(\frac{x+1}{x+3}\right)}^2$