Rút gọn các phân thức Bài 9 trang 26 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 3: Rút gọn phân thức

Bài 9 trang 26 SBT Toán 8 Tập 1: 

a) $\frac{14x{y}^5(2x-3y)}{21x^{2}y{(2x-3y)}^2}$;

b) $\frac{8xy{(3x-1)}^3}{12x^3(1-3x)}$;

c) $\frac{20x^2-45}{{(2x+\text{\hspace{0.17em}}3)}^2}$;

d) $\frac{5x^2-10xy}{2{(2y-x)}^3}$;

e) $\frac{80x^3-125x}{3(x-3)-(x-3).(8-4x)}$;

f) $\frac{9-{(x+\text{\hspace{0.17em}}5)}^2}{x^2+4x+\text{\hspace{0.17em}}4}$;

g) $\frac{32x-8x^2+\text{\hspace{0.17em}}2x^3}{x^3+\text{\hspace{0.17em}}64}$;

h) $\frac{5x^3+\text{\hspace{0.17em}}5x}{x^4-1}$;

i) $\frac{x^2+\text{\hspace{0.17em}\hspace{0.17em}}5x+\text{\hspace{0.17em}}6}{x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}4}$.

Lời giải:

a) $\frac{14x{y}^5(2x-3y)}{21x^{2}y{(2x-3y)}^2}=\frac{2y^4}{3x(2x-3y)}$;

b)

$\frac{8xy{(3x-1)}^3}{12x^3(1-3x)}=\frac{2y{(3x-1)}^3}{3x^2.\left.-(3x-1)\right]}=-\frac{2y{(3x-1)}^2}{3x^2}$

c)

$\frac{20x^2-45}{{(2x+\text{\hspace{0.17em}}3)}^2}=\frac{5(4x^2-9)}{{(2x+\text{\hspace{0.17em}}3)}^2}=\frac{5(2x+3)(2x-3)}{{(2x+\text{\hspace{0.17em}}3)}^2}=\frac{5(2x-3)}{2x+3}$

d)

$\frac{5x^2-10xy}{2{(2y-x)}^3}\text{\hspace{0.17em}\hspace{0.17em}}=\frac{5x(x-2y)}{2{(2y-x)}^3}=\frac{-5x(2y-x)}{2{(2y-x)}^3}=\frac{-5x}{2{(2y-x)}^2}$

e)

$\frac{80x^3-125x}{3(x-3)-(x-3).(8-4x)}=\frac{5x(16x^2-25)}{(x-3).(3-8+\text{\hspace{0.17em}}4x)}=\frac{5x(4x+\text{\hspace{0.17em}}5)(4x-5)}{(x-3).(4x-5)}=\frac{5x(4x+\text{\hspace{0.17em}\hspace{0.17em}}5)}{x-3}$

f)

$\begin{array}{l}\frac{9-{(x+\text{\hspace{0.17em}}5)}^2}{x^2+4x+\text{\hspace{0.17em}}4}\text{\hspace{0.17em}}=\frac{\left.3+(x+\text{\hspace{0.17em}}5)\right].\left.3-(x+5)\right]}{{(x+\text{\hspace{0.17em}}2)}^2}\\ =\frac{(8+\text{\hspace{0.17em}}x).(-x-2)}{{(x+\text{\hspace{0.17em}}2)}^2}=\frac{-(8+\text{\hspace{0.17em}}x).(x+2)}{{(x+\text{\hspace{0.17em}}2)}^2}\\ =\frac{-(8+\text{\hspace{0.17em}}x)}{x+\text{\hspace{0.17em}}2}\end{array}$

g)

$\frac{32x-8x^2+\text{\hspace{0.17em}}2x^3}{x^3+\text{\hspace{0.17em}}64}=\frac{2x(16-4x\text{\hspace{0.17em}\hspace{0.17em}}+x^2)}{(x+\text{\hspace{0.17em}}4).(x^2-4x+16)}=\frac{2x}{x+\text{\hspace{0.17em}}4}$

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

i)

$\begin{array}{l}\frac{x^2+\text{\hspace{0.17em}\hspace{0.17em}}5x+\text{\hspace{0.17em}}6}{x^2+\text{\hspace{0.17em}\hspace{0.17em}}4x+\text{\hspace{0.17em}}4}=\frac{x^2+2x+3x+6}{{(x+2)}^2}\\ =\frac{x(x+2)\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}\hspace{0.17em}}3(x+2)}{{(x+\text{}2)}^2}\\ =\frac{(x+3).(x+\text{\hspace{0.17em}\hspace{0.17em}}2)}{{(x+2)}^2}=\frac{x+3}{x+2}\end{array}$