Thực hiện phép chia phân thức Bài 39 trang 34 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 8: Phép chia các phân thức đại số

Bài 39 trang 34 SBT Toán 8 Tập 1: Thực hiện phép chia phân thức:

a) $\frac{x^2-5x+\text{\hspace{0.17em}}6}{x^2\text{}+\text{\hspace{0.17em}}7x+12}:\frac{x^2-4x+4}{x^2+3x}$;

b) $\frac{x^2+\text{}2x-3}{x^2+\text{\hspace{0.17em}}3x-10}:\text{\hspace{0.17em}\hspace{0.17em}}\frac{x^2+\text{\hspace{0.17em}}7x+12}{x^2-9x+14}$.

Lời giải:

a) $\frac{x^2-5x+\text{\hspace{0.17em}}6}{x^2\text{}+\text{\hspace{0.17em}}7x+12}:\frac{x^2-4x+4}{x^2+3x}$

$=\frac{x^2-5x+\text{\hspace{0.17em}}6}{x^2\text{}+\text{\hspace{0.17em}}7x+12}.\frac{x^2+3x}{x^2-4x+4}$

$=\frac{(x^2-5x+\text{\hspace{0.17em}}6)(x^2+3x)}{(x^2\text{}+\text{\hspace{0.17em}}7x+12)(x^2-4x+4)}$

$=\frac{\text{[}(x^2-2x)-(3x-\text{\hspace{0.17em}}6)\text{]x}(x+3)}{\text{[}(x^2\text{}+\text{\hspace{0.17em}}3x)+(4x+12)\text{]}{(x-2)}^2}$

$=\frac{\text{[x}(x-2)-3(x-\text{\hspace{0.17em}}2)\text{]x}(x+3)}{\text{[x}(x\text{}+\text{\hspace{0.17em}}3)+4(x+3)\text{]}{(x-2)}^2}$

$=\frac{(x-2)(x-3)x(x+\text{}3)}{(x+4)(x+3){(x-2)}^2}=\frac{x(x-3)}{(x+\text{}4)(x-2)}$

b) $\frac{x^2+\text{}2x-3}{x^2+\text{\hspace{0.17em}}3x-10}:\text{\hspace{0.17em}\hspace{0.17em}}\frac{x^2+\text{\hspace{0.17em}}7x+12}{x^2-9x+14}$

$=\frac{x^2+\text{}2x-3}{x^2+\text{\hspace{0.17em}}3x-10}.\frac{x^2-\text{\hspace{0.17em}}9x+14}{x^2+7x+12}$

$=\frac{(x^2+\text{}2x-3)(x^2-9x+14)}{(x^2+\text{\hspace{0.17em}}3x-10)(x^2+7x+12)}$

$\text{=\hspace{0.17em}}\frac{\left.(x^2+3x)-(x+\text{}3)\right].\left.(x^2-2x)-(7x-14)\right]}{\left.(x^2-2x)+(5x-10)\right]\left.(x^2+3x)+\text{}(4x+\text{}12)\right]}$

$=\frac{\left.x(x+3)-1.(x+\text{}3)\right].\left.x(x-2)-7(x-2)\right]}{\left.x(x-2)+5(x-2)\right]\left.x(x+3)+4\text{}(x+\text{}3)\right]}$

$=\frac{(x-1).(x+\text{}3)(x-7)(x-2)}{(x+\text{\hspace{0.17em}}5)(x-2)(x+\text{}4)(x+3)}=\frac{(x-1)(x-7)}{(x+\text{\hspace{0.17em}}5)(x+4)}$