Làm tính nhân phân thức Bài 29 trang 32 SBT Toán 8 Tập 1

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

Bài 29 trang 32 SBT Toán 8 Tập 1: Làm tính nhân phân thức:

a) $\frac{30x^3}{11y^2}.\frac{121y^5}{25x}$;

b) $\frac{24y^5}{7x^2}.\left(\frac{-21x}{12y^3}\right)$;

c) $\left(\frac{-18y^3}{25x^4}\right).\left(\frac{-15x^2}{9y^3}\right)$;

d) $\frac{4x+8}{{(x-10)}^3}.\frac{2x-20}{{(x+\text{\hspace{0.17em}}2)}^2}$;

e) $\frac{2x^2-20x\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}50}{3x+\text{\hspace{0.17em}\hspace{0.17em}}3}.\frac{x^2-1}{4{(x-5)}^3}$.

Lời giải:

a) 

$\begin{array}{l}\frac{30x^3}{11y^2}.\frac{121y^5}{25x}=\frac{30x^3.121y^5}{11y^2.25x}\\ =\frac{6x^2.11y^3}{5}=\frac{66x^2{y}^3}{5}\end{array}$

b) 

$\begin{array}{l}\frac{24y^5}{7x^2}.\left(\frac{-21x}{12y^3}\right)=\frac{24y^5.(-21x)}{7x^2.12y^3}\\ =\frac{2y^2.(-3)}{x}=\frac{-6y^2}{x}\end{array}$

c) 

$\begin{array}{l}\left(\frac{-18y^3}{25x^4}\right).\left(\frac{-15x^2}{9y^3}\right)=\frac{-18y^3.(-15x^2)}{25x^4.9y^3}\\ =\frac{2.3}{5x^2}=\frac{6}{5x^2}\end{array}$

d)

$\frac{4x+8}{{(x-10)}^3}.\frac{2x-20}{{(x+\text{\hspace{0.17em}}2)}^2}=\frac{(4x+8)(2x-20)}{{(x-10)}^3.{(x+\text{}2)}^2}$

$=\frac{4(x+2).2(x-10)}{{(x-10)}^3{(x+\text{}2)}^2}=\frac{8}{{(x-10)}^2(x+2)}$

e)

.$\frac{2x^2-20x\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}50}{3x+\text{\hspace{0.17em}\hspace{0.17em}}3}.\frac{x^2-1}{4{(x-5)}^3}=\frac{(2x^2-20x\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}50)(x^2-1)}{(3x+\text{\hspace{0.17em}\hspace{0.17em}}3)4.{(x-5)}^3}$

$=\frac{2(x^2-10x\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}25)(x+1)(x-1)}{3(x+\text{\hspace{0.17em}}1)4.{(x-5)}^3}=\frac{2{(x-5)}^2(x+1)(x-1)}{12(x+\text{\hspace{0.17em}}1).{(x-5)}^3}$

$=\frac{x-1}{6(x-5)}$