Tìm phân thức P Bài 8.2 trang 35 SBT Toán 8 Tập 1

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

Bài 8.2 trang 35 SBT Toán 8 Tập 1: Tìm phân thức P biết :

a) $P:\text{\hspace{0.17em}\hspace{0.17em}}\frac{4x^2-\text{\hspace{0.17em}}16}{2x+1}=\frac{4x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}1}{x-2}$;

b) $\frac{2x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}8}{x^3-3x^2-x+\text{\hspace{0.17em}}3}:\text{\hspace{0.17em}}P=\frac{x^3-8}{(x+\text{\hspace{0.17em}}1).(x-3)}$.

Lời giải:

a) $P:\text{\hspace{0.17em}\hspace{0.17em}}\frac{4x^2-\text{\hspace{0.17em}}16}{2x+1}=\frac{4x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}1}{x-2}$

$\begin{array}{l}P=\frac{4x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}1}{x-2}.\text{}\frac{4x^2-\text{\hspace{0.17em}}16}{2x+1}\\ P=\frac{(4x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}1)(4x^2-16)}{(x-2)(2x+1)}\end{array}$

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

$P=\frac{{(2x+1)}^2.2(x+2)2(x-2)}{(x-2)(2x+1)}$

$P=4(2x+1).(x+\text{}2)$

Vậy $P=4(2x+1).(x+\text{}2)$.

b)$\frac{2x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}8}{x^3-3x^2-x+\text{\hspace{0.17em}}3}:\text{\hspace{0.17em}}P=\frac{x^3-8}{(x+\text{\hspace{0.17em}}1).(x-3)}$

$\begin{array}{l}P=\frac{2x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}8}{x^3-3x^2-x+\text{\hspace{0.17em}}3}:\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\frac{x^3-8}{(x+\text{\hspace{0.17em}}1).(x-3)}\\ P\text{}=\frac{2(x^2+\text{\hspace{0.17em}\hspace{0.17em}}2x+\text{\hspace{0.17em}}4)}{x^2(x-3)-(x-\text{\hspace{0.17em}}3)}.\frac{(x+\text{}1).(x-3)}{x^3-8}\end{array}$

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

$\begin{array}{l}=\frac{2(x^2+\text{\hspace{0.17em}\hspace{0.17em}}2x+\text{\hspace{0.17em}}4).(x+1)(x-3)}{(x+1)(x-1)(x-3)(x-2)(x^2+\text{}2x+4)}\\ =\frac{2}{(x-1)(x-2)}\end{array}$