Bài tập về cộng và trừ các phân thức đại số
1. Thực hiện các phép tính:
a) $\frac{x}{x+1}-\frac{x^{3}-2x^{2}}{x^{3}+1}$
b) $\frac{x+1}{2x-2}+\frac{3}{x^{2}-1}-\frac{x+3}{2x+2}$
c) $\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x(1-x)}{9-x^{2}}$
d) $\frac{3x+1}{(x-1)^{2}}-\frac{1}{x+1}+\frac{x+3}{1-x^{2}}$
2. a) Rút gọn biểu thức: $\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^{2}}+\frac{4}{1+x^{4}}+\frac{8}{1+x^{8}}$
b) Xác định a, b để: $\frac{1}{x^{2}-4}=\frac{a}{x-2}+\frac{b}{x+2}$
3. a) Xác định a, b để: $\frac{1}{x(x-1)}=\frac{a}{x-1}+\frac{b}{x}$
b) Áp dụng, rút gọn biểu thức:
M = $\frac{1}{x^{2}-5x+6}+\frac{1}{x^{2}-7x+12}+\frac{1}{x^{2}-9x+20}+\frac{1}{x^{2}-11x+30}$
4. Xác định a, b, c để ta có: $\frac{1}{x(x^{2}+1)}=\frac{a}{x}+\frac{bx+c}{x^{2}+1}$
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 $\frac{1}{x(x^{2}+1)}=\frac{1}{x}-\frac{x}{x^{2}+1}$
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