Rút gọn các biểu thức Bài 41 trang 34 SBT Toán 8 Tập 1

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

Bài 41 trang 34 SBT Toán 8 Tập 1: 

a) $\frac{x+1}{x+2}:\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}$;

b) $\frac{x+1}{x+2}:\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}\right)$;

c) $\frac{x+1}{x+2}.\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}$;

d) $\frac{x+1}{x+2}.\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}\right)$;

e) $\frac{x+1}{x+2}:\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+3}{x+\text{\hspace{0.17em}}1}$;

f) $\frac{x+1}{x+2}:\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+3}{x+\text{\hspace{0.17em}}1}\right)$.

Lời giải:

a) $\frac{x+1}{x+2}:\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}$

$=\frac{x+1}{x+2}.\frac{x+\text{\hspace{0.17em}}3}{x+2}.\frac{x+1}{x+\text{\hspace{0.17em}}3}$

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

b) $\frac{x+1}{x+2}:\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}\right)$

$\begin{array}{l}=\frac{x+1}{x+2}:\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+1}{x+\text{\hspace{0.17em}}3}\right)\\ =\frac{x+1}{x+2}:\frac{(x+\text{\hspace{0.17em}}2)(x+\text{\hspace{0.17em}}1)}{{(x+3)}^2}\end{array}$

$=\frac{x+\text{\hspace{0.17em}}1}{x+2}.\frac{{(x+\text{\hspace{0.17em}}3)}^2}{(x+\text{}2).(x+1)}$

$=\frac{(x+\text{\hspace{0.17em}}1){(x+\text{\hspace{0.17em}}3)}^2}{(x+2)(x+\text{}2).(x+1)}=\frac{{(x+\text{\hspace{0.17em}}3)}^2}{{(x+2)}^2}$

c) $\frac{x+1}{x+2}.\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}$

$=\frac{x+1}{x+2}.\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+1}{x+\text{\hspace{0.17em}}3}$

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

d) $\frac{x+1}{x+2}.\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}:\frac{x+3}{x+\text{\hspace{0.17em}}1}\right)$

$=\frac{x+1}{x+2}.\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+1}{x+\text{\hspace{0.17em}}3}\right)$

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

e) $\frac{x+1}{x+2}:\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+3}{x+\text{\hspace{0.17em}}1}$

$=\frac{x+1}{x+2}.\frac{x+\text{\hspace{0.17em}}3}{x+2}.\frac{x+3}{x+\text{\hspace{0.17em}}1}$

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

f) $\frac{x+1}{x+2}:\left(\frac{x+\text{\hspace{0.17em}}2}{x+3}.\frac{x+3}{x+\text{\hspace{0.17em}}1}\right)$

$=\frac{x+1}{x+2}:\frac{x+2}{x+1}$

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