Bài tập về cộng và trừ các phân thức đại số

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 $\{\begin{array}{c}a+b=0\\ 2a-2b=1\end{array}\iff \{\begin{array}{c}x=\frac{1}{4}\\ b=-\frac{1}{4}\end{array}$

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 $\{\begin{array}{c}a+b=0\\ b=-1\end{array}$

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: $\{\begin{array}{c}a+b=0\\ c=0\\ a=1\end{array}$

Ta tính được a=1; b=-1; c=0

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