Question

bài1: thực hiện phép tính

a,\(\frac{1}{\left(1-x\right).\left(2-x\right)}+\frac{2}{\left(2-x\right).\left(3-x\right)}+\frac{3}{\left(1-x\right).\left(x-3\right)}\)

b,\(\frac{x^2}{x+1}+\frac{2x}{^{ }x^2-1}-\frac{1}{1-x}+1\)

c,\(\frac{1}{x^3-x}-\frac{1}{\left(x-1\right).x}+\frac{2}{x^2-1}\)

Answer 1

Answer 2

Lời giải:

a)

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

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

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

b)

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

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

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

c)

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

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