Rút gọn biểu thức Bài 26 trang 31 SBT Toán 8 Tập 1

Giải SBT Toán 8 Bài 6: Phép trừ các phân thức đại số

Bài 26 trang 31 SBT Toán 8 Tập 1: Rút gọn biểu thức:

a) $\frac{3x^2+5x+\text{\hspace{0.17em}}1}{x^3-1}-\frac{1-x}{x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+\text{\hspace{0.17em}}1}-\frac{3}{x-1}$;

b) $\frac{1}{x^2-x+\text{\hspace{0.17em}}1}+1-\frac{x^2+\text{\hspace{0.17em}}2}{x^3+\text{\hspace{0.17em}}1}$;

c) $\frac{7}{x}-\frac{x}{x+\text{\hspace{0.17em}\hspace{0.17em}}6}+\frac{36}{x^2+\text{\hspace{0.17em}\hspace{0.17em}}6x}$.

Lời giải:

a) $\frac{3x^2+5x+\text{\hspace{0.17em}}1}{x^3-1}-\frac{1-x}{x^2+\text{\hspace{0.17em}\hspace{0.17em}}x+\text{\hspace{0.17em}}1}-\frac{3}{x-1}$

b) $\frac{1}{x^2-x+\text{\hspace{0.17em}}1}+1-\frac{x^2+\text{\hspace{0.17em}}2}{x^3+\text{\hspace{0.17em}}1}$

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

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

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

c) $\frac{7}{x}-\frac{x}{x+\text{\hspace{0.17em}\hspace{0.17em}}6}+\frac{36}{x^2+\text{\hspace{0.17em}\hspace{0.17em}}6x}$

$=\frac{7}{x}-\frac{x}{x+\text{\hspace{0.17em}\hspace{0.17em}}6}+\frac{36}{x(x+\text{\hspace{0.17em}\hspace{0.17em}}6)}$

$=\frac{7(x+6)-x.x+36}{x(x+\text{\hspace{0.17em}\hspace{0.17em}}6)}=\frac{7x+42-x^2+36}{x(x+\text{\hspace{0.17em}\hspace{0.17em}}6)}$

$=\frac{-x^2+7x+78}{x(x+6)}=\frac{-x^2-6x+13x+78}{x(x+6)}$

$=\frac{-x(x+6)+13(x+6)}{x(x+6)}$

$=\frac{(x+6)(-x+13)}{x(x+6)}=\frac{-x+13}{x}$