Trong mỗi trường hợp sau hãy tìm phân thức Q thỏa mãn điều kiện

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

Bài 6.2 trang 32 SBT Toán 8 Tập 1: 

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

b) $\frac{2x-6}{x^3-3x^2-x+3}\text{\hspace{0.17em}\hspace{0.17em}}+\text{\hspace{0.17em}}Q=\frac{6}{x-3}-\frac{2x^2}{1-x^2}$.

Lời giải:

a) Ta có:

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

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

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

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

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

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

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

$\begin{array}{l}=\frac{6x^2-6+2x^3-6x^2-2x+6}{(x^2-1)(x-3)}\\ =\frac{2x^3-2x}{(x^2-1)(x-3)}=\frac{2x(x^2-1)}{(x^2-1)(x-3)}=\frac{2x}{x-3}\end{array}$