Bài tập về cộng và trừ các phân thức đại số

1. 

a) $\frac{x}{x+1}-\frac{x^{3}-2x^{2}}{x^{3}+1}$

 = $\frac{x(x^{2}-x+1)}{(x+1)(x^{2}-x+1)}-\frac{x^{3}-2x^{2}}{(x+1)(x^{2}-x+1)}$

 = $\frac{x^{3}-x^{2}+x-x^{3}+2x^{2}}{(x+1)(x^{2}-x+1)}$

 = $\frac{x^{2}+x}{(x+1)(x^{2}-x+1)}$

 = $\frac{x(x+1)}{(x+1)(x^{2}-x+1)}$ 

 = $\frac{x}{x^{2}-x+1}$

b) $\frac{x+1}{2x-2}+\frac{3}{x^{2}-1}-\frac{x+3}{2x+2}$

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

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

 = $\frac{10}{2(x-1)(x+1)}$

 = $\frac{5}{(x-1)(x+1)}$

c) $\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x(1-x)}{9-x^{2}}$

 = $\frac{(x+1)(x+3)}{(x-3)(x+3)}-\frac{(1-x)(x-3)}{(x-3)(x+3)}-\frac{2x(1-x)}{(x-3)(x+3)}$

 = $\frac{(x+1)(x+3)-(1-x)(x-3)+2x(1-x)}{(x-3)(x+3)}$

 = $\frac{2x+6}{(x-3)(x+3)}$

 = $\frac{2}{x-3}$

d) $\frac{3x+1}{(x-1)^{2}}-\frac{1}{x+1}+\frac{x+3}{1-x^{2}}$

 = $\frac{(3x+1)(x+1)}{(x-1)^{2}(x+1)}-\frac{x^{2}-1}{(x-1)^{2}(x+1)}-\frac{(x-1)(x+3)}{(x-1)^{2}(x+1)}$

 = $\frac{x^{2}+4x+3}{(x-1)^{2}(x+1)}$

 = $\frac{(x+1)(x+3)}{(x-1)^{2}(x+1)}$

 = $\frac{x+3}{(x-1)^{2}}$

2. a) $\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}$

    = $\frac{1+x+1-x}{1-x^{2}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}$

    = $\frac{2}{1-x^{2}}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}$

    = $\frac{2(1+x^{2})+2(1-x^{2})}{1-x^{4}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}$

    = $\frac{4}{1-x^{4}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}$

    = $\frac{4(1+x^{4})+4(1-x^{4})}{1-x^{8}}+\frac{8}{1+x^{8}}$

    = $\frac{8}{1-x^{8}}+\frac{8}{1+x^{8}}$

    = $\frac{8(1+x^{8})+8(1-x^{8})}{1-x^{16}}$

    = $\frac{16}{1-x^{16}}$

   b) Ta có: $\frac{1}{x^{2}-4}=\frac{a}{x-2}+\frac{b}{x+2}$

                               = $\frac{(a+b)x+2(a-b)}{x^{2}-4}$

Từ đó ta có $\left\{\begin{matrix}a+b=0\\ 2a-2b=1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x=\frac{1}{4}\\ b=-\frac{1}{4}\end{matrix}\right.$

3. a) $\frac{1}{x(x-1)}=\frac{a}{x-1}+\frac{b}{x}=\frac{ax+b(x-1)}{x(x-1)}=\frac{ax+bx-b}{x(x-1)}=\frac{x(a+b)-b}{x(x-1)}$

Từ đó ta có $\left\{\begin{matrix}a+b=0\\ b=-1\end{matrix}\right.$

Ta được a = 1; b = -1

   b) Áp dụng ta có: 

M = $\frac{1}{x^{2}-5x+6}+\frac{1}{x^{2}-7x+12}+\frac{1}{x^{2}-9x+20}+\frac{1}{x^{2}-11x+30}$

  = $\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}+\frac{1}{(x-4)(x-5)}+\frac{1}{(x-5)(x-6)}$

  = $\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-6}-\frac{1}{x-5}$

  = $\frac{1}{x-6}-\frac{1}{x-2}$

  = $\frac{4}{(x-2)(x-6)}$

4. Ta có: $\frac{1}{x(x^{2}+1)}=\frac{a}{x}+\frac{bx+c}{x^{2}+1}=\frac{(a+b)x^{2}+cx+a}{x(x^{2}+1)}$

Từ đó ta có: $\left\{\begin{matrix}a+b=0\\ c=0\\ a=1\end{matrix}\right.$

Ta tính được a=1; b=-1; c=0

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