Rút gọn các biểu thức Bài 30 trang 32 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 7. Phép nhân các phân thức đại số

Bài 30 trang 32 SBT Toán 8 Tập 1: Rút gọn các biểu thức (chú ý dùng quy tắc đổi dấu để thay nhân tử chung).

a) $\frac{x+3}{x^2-4}.\frac{8-12x+\text{\hspace{0.17em}\hspace{0.17em}}6x^2-x^3}{9x+\text{\hspace{0.17em}\hspace{0.17em}}27}$;

b) $\frac{6x-3}{5x^2+x}.\frac{25x^2\text{}+\text{\hspace{0.17em}\hspace{0.17em}}10x+\text{\hspace{0.17em}}1}{1-8x^3}$;

c) $\frac{3x^2-x}{x^2-1}.\frac{1-x^4}{{(1-3x)}^3}$.

Lời giải:

a) $\frac{x+3}{x^2-4}.\frac{8-12x+\text{\hspace{0.17em}\hspace{0.17em}}6x^2-x^3}{9x+\text{\hspace{0.17em}\hspace{0.17em}}27}$

$=\frac{x+3}{(x+2)(x-2)}.\frac{-(x^3-6x^2+12x-8)}{9(x+\text{\hspace{0.17em}\hspace{0.17em}}3)}$

$=\frac{x+3}{(x+2)(x-2)}.\frac{-{(x-2)}^3}{9(x+\text{\hspace{0.17em}\hspace{0.17em}}3)}$

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

b) $\frac{6x-3}{5x^2+x}.\frac{25x^2\text{}+\text{\hspace{0.17em}\hspace{0.17em}}10x+\text{\hspace{0.17em}}1}{1-8x^3}$

$=\frac{3(2x-1)}{x(5x+1)}.\frac{{(5x+1)}^2}{-(8x^3-1)}$

$=\frac{3(2x-1).{(5x+1)}^2}{-x(5x+1)(2x-1).(4x^2-2x+1)}$

$=\frac{3(5x+\text{}1)}{-x(4x^2-2x+1)}=\frac{-3(5x+\text{}1)}{x(4x^2-2x+1)}$

c) $\frac{3x^2-x}{x^2-1}.\frac{1-x^4}{{(1-3x)}^3}$

$=\frac{x(3x-1)}{(x+1)(x-1)}.\frac{(1-x^2)(1+\text{}{x}^2)}{-{(3x-1)}^3}$

$=\frac{x(3x-1)}{(x+1)(x-1)}.\frac{-(x^2-1)(1+\text{}{x}^2)}{-{(3x-1)}^3}$

$\begin{array}{l}=\frac{x(3x-1)}{(x+1)(x-1)}.\frac{(x+1)(x-1)(1+\text{}{x}^2)}{{(3x-1)}^3}\\ =\frac{x(3x-1).(x+1)(x-1)(1+\text{}{x}^2)}{(x+1)(x-1){(3x-1)}^3}\\ =\frac{x(1+\text{}x)_2}{{(3x-1)}^2}\end{array}$