Lời giải:

a. ĐKXĐ: $x\neq 0;-1$

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

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

b. ĐKXĐ: $x\neq 0; 1;2$

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

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

c. ĐKXĐ: $x\neq 0; -1$
\(=\frac{x+1+x^2}{x(x+1)}.\frac{x(x+1)}{x}=\frac{x^2+x+1}{x}\)

b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)

\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)

\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)

\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)

\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)

c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)

\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)

\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)

\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)

rảnh vãi

thanks nha 

a: \(\frac{2}{x+2}-\frac{1}{x+3}+\frac{2x+5}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{2\left(x+3\right)-\left(x+2\right)+2x+5}{\left(x+2\right)\left(x+3\right)}=\frac{2x+6+2x+5-x-2}{\left(x+2\right)\left(x+3\right)}\)

\(=\frac{3x+9}{\left(x+2\right)\left(x+3\right)}=\frac{3\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}=\frac{3}{x+2}\)

b: \(\frac{2}{x+1}-\frac{1}{x+5}+\frac{2x+6}{\left(x+5\right)\left(x+1\right)}\)

\(=\frac{2\left(x+5\right)-x-1+2x+6}{\left(x+1\right)\left(x+5\right)}\)

\(=\frac{2x+10-x-1+2x+6}{\left(x+1\right)\left(x+5\right)}=\frac{3x+15}{\left(x+1\right)\left(x+5\right)}\)

\(=\frac{3\left(x+5\right)}{\left(x+1\right)\left(x+5\right)}=\frac{3}{x+1}\)

c: \(\frac{-6}{x^2-9}-\frac{1}{x+3}+\frac{3}{x-3}\)

\(=\frac{-6}{\left.\left(x-3\right)\left(x+3\right)\right.}-\frac{1}{x+3}+\frac{3}{x-3}\)

\(=\frac{-6-\left(x-3\right)+3\left(x+3\right)}{\left.\left(x-3\right)\left(x+3\right)\right.}=\frac{-6-x+3+3x+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\frac{2x+6}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{x-3}\)

d: \(\frac{x}{x-2}-\frac{x}{x+2}+\frac{8}{x^2-4}\)

\(=\frac{x}{x-2}-\frac{x}{x+2}+\frac{8}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x\left(x+2\right)-x\left(x-2\right)+8}{\left(x-2\right)\left(x+2\right)}=\frac{x^2+2x-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{4x+8}{\left(x-2\right)\left(x+2\right)}=\frac{4}{x-2}\)

help me pls!!!

giúp bạn cx hơi hảo tổn đó :))

\(a,=\left(\dfrac{1-x}{x}+\dfrac{x^3-x}{x}\right)\times\dfrac{x}{x-1}\\ =\dfrac{1-x+x^3-x}{x}\times\dfrac{x}{x-1}\\ =\dfrac{1-2x+x^3}{x-1}\\ b,=\left(\dfrac{x-x^2}{x.x^2}\right).\dfrac{x^2}{y}+\dfrac{x}{y}\\ =\dfrac{x-x^2}{xy}+\dfrac{x}{y}\\ =\dfrac{x-x^2+x^2}{xy}=\dfrac{x}{xy}=\dfrac{1}{y}\)

\(c,=\dfrac{3}{x}-\dfrac{2}{x}\times x+\dfrac{x}{3}\\ =\dfrac{3}{x}-2+\dfrac{x}{3}\\ =\dfrac{3-2x+x^2}{3x}\)

a)

\(\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{x+1-x}{x\left(x+1\right)}=\dfrac{1}{x\left(x+1\right)}\left(đpcm\right)\)

b)

\(\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}\\ =\dfrac{1}{x}\)