Question

Answer 1

a/ ĐKXĐ: $x\neq \pm 1$
\(\frac{1}{x+1}+\frac{1}{1-x}+\frac{2x^2}{x^2-1}\\ =\frac{x-1}{(x+1)(x-1)}-\frac{x+1}{(x-1)(x+1)}+\frac{2x^2}{(x-1)(x+1)}\\ =\frac{x-1-(x+1)+2x^2}{(x-1)(x+1)}=\frac{2x^2-2}{(x-1)(x+1)}\\ =\frac{2(x^2-1)}{x^2-1}=2\)

b/ ĐKXĐ: $x\neq \pm 2$

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

 

Answer 2

c/ ĐKXĐ: $x\neq 1$

\(=\frac{4x^2-3x+17}{(x-1)(x^2+x+1)}+\frac{(2x-1)(x-1)}{(x-1)(x^2+x+1)}-\frac{6(x^2+x+1)}{(x-1)(x^2+x+1)}\\ =\frac{4x^2-3x+17+(2x^2-3x+1)-(6x^2+6x+6)}{(x-1)(x^2+x+1)}\)

\(=\frac{-12x+12}{(x-1)(x^2+x+1)}=\frac{-12(x-1)}{(x-1)(x^2+x+1)}=\frac{-12}{x^2+x+1}\)

Answer 3

d/ ĐKXĐ: $x\neq -1$
\(=\frac{x^2-x+1}{(x+1)(x^2-x+1)}-\frac{3}{(x+1)(x^2-x+1)}+\frac{3(x+1)}{(x+1)(x^2-x+1)}\\ =\frac{x^2-x+1-3+3x+3}{(x+1)(x^2-x+1)}\\ =\frac{x^2+2x+1}{(x+1)(x^2-x+1)}=\frac{(x+1)^2}{(x+1)(x^2-x+1)}\\ =\frac{x+1}{x^2-x+1}\)