Phân tích các mẫu thức và các mẫu thức rồi rút gọn biểu thức

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

Bài 31 trang 32 SBT Toán 8 Tập 1: 

a) $\frac{x-2}{x+1}.\frac{x^2-2x-3}{x^2-5x+6}$;

b) $\frac{x\text{\hspace{0.17em}}+1}{x^2-2x-8}.\text{}\frac{4-x}{x^2\text{}+x}$;

c) $\frac{x+\text{\hspace{0.17em}}2}{4x\text{\hspace{0.17em}}+24}.\frac{x^2-36}{x^2\text{}+x-2}$.

Lời giải:

a) $\frac{x-2}{x+1}.\frac{x^2-2x-3}{x^2-5x+6}$

$\begin{array}{l}=\frac{(x-2)(x^2-2x-3)}{(x+\text{}1)(x^2-5x+6)}\\ =\frac{(x-2)(x^2+x-3x-3)}{(x+\text{}1)(x^2-2x-3x+6)}\end{array}$

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

b) $\frac{x\text{\hspace{0.17em}}+1}{x^2-2x-8}.\text{}\frac{4-x}{x^2\text{}+x}$

$\begin{array}{l}=\frac{(x\text{\hspace{0.17em}}+1)(4-x)}{(x^2-2x-8)(x^2+x)}\\ =\frac{-(x+\text{\hspace{0.17em}}1).(x-4)}{\left.(x^2-4x)+(2x-8)\right]\left(x+_2\text{}x\right)}\end{array}$

$\begin{array}{l}=\frac{-(x+\text{}1).(x-4)}{\text{[}x(x-4)+\text{}2(x-4)\text{].x(x\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}1)}}\\ =\frac{-(x+\text{}1).(x-4)}{(x+2)(x-4)\text{x(x\hspace{0.17em}\hspace{0.17em}+\hspace{0.17em}1)}}=\frac{-1}{x(x+\text{}2)}\end{array}$

c) $\frac{x+\text{\hspace{0.17em}}2}{4x\text{\hspace{0.17em}}+24}.\frac{x^2-36}{x^2\text{}+x-2}$

$=\frac{(x+\text{}2).(x^2-36)}{(4x+24)(x^2\text{}+x-2)}$

$=\frac{(x+\text{}2).(x+\text{}6)(x-6)}{4(x+6)(x^2\text{}+2x-x-2)}$

$=\frac{(x+\text{}2).(x+\text{}6)(x-6)}{4(x+6)\text{[x}(x\text{}+2)-(x+2)\text{]}}$

$=\frac{(x+\text{}2).(x+\text{}6)(x-6)}{4(x+6).(x-1).(x+2)}=\text{\hspace{0.17em}\hspace{0.17em}}\frac{x-6}{4(x-1)}$