a:

ĐKXĐ: x<>-1

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

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

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

b: \(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\)

\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)

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

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

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

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

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

\(=\dfrac{x^2-4x-x^2+4}{x\left(x^2-4\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)

c: \(\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\)

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

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

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

d:

\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)

\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

đk: \(_{x+1\ne0\Leftrightarrow x\ne-1}\)\(\dfrac{1-x}{x+1}+3=\dfrac{2x-3}{x+1}\Leftrightarrow\dfrac{1-x}{x+1}+\dfrac{3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x-1}\Leftrightarrow1-x+3x+3-2x-3=0\Leftrightarrow-2x+1=0\Leftrightarrow-2x=-1\Leftrightarrow x=0,5\)

a) Ta có: \(\dfrac{x}{x-1}-\dfrac{2}{x-1}\)

\(=\dfrac{x-2}{x-1}\)

b) Ta có: \(\dfrac{4+4x}{3x^2+6x}+\dfrac{x}{3x+6}\)

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

\(=\dfrac{x^2+4x+4}{3x\left(x+2\right)}\)

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

\(=\dfrac{x+2}{3x}\)

c) Ta có: \(\dfrac{x^2-2x}{x-1}\cdot\dfrac{1}{x}:\dfrac{x^2-4}{x^2-2x+1}\)

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

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

\(=\dfrac{x-1}{x+2}\)

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

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

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

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

\(a,\dfrac{1}{x+2}=\dfrac{x\left(2-x\right)}{x\left(2-x\right)\left(2+x\right)}\\ \dfrac{8}{2x-x^2}=\dfrac{8}{x\left(2-x\right)}=\dfrac{8\left(2+x\right)}{x\left(2-x\right)\left(2+x\right)}\)

\(b,\dfrac{2-x}{x^2-9}=\dfrac{2-x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x\left(2-x\right)}{x\left(x-3\right)\left(x+3\right)}\\ \dfrac{-1}{x^2+3x}=\dfrac{-1}{x\left(x+3\right)}=\dfrac{-\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3-x}{x\left(x-3\right)\left(x+3\right)}\)

a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)

\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)

\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)

\(\Leftrightarrow1=0\)(vô lý)

Vậy: \(S=\varnothing\)

Ai giúp vs

\(a,\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\left(ĐKXĐ:x\ne-1\right)\\ =\dfrac{x^2+2-\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{1}{x^2-x+1}\\ c,\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\\ =\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm1\right)\\ =\dfrac{-1\left(x+1\right)-3\left(x-1\right)+2x.2}{2\left(x+1\right)\left(x-1\right)}\\ =\dfrac{-x-1-3x+3+4x}{2\left(x+1\right)\left(x-1\right)}=\dfrac{2}{2\left(x+1\right)\left(x-1\right)}=\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

\(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\) (ĐK: \(x\ne0;x\ne\pm2\) )
\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)

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

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

\(=\dfrac{x^2+2x-x^2-4x-2x+4}{x\left(x+2\right)\left(x-2\right)}\)

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

\(=\dfrac{4-4x}{x^3-4x}\) 

\(b,\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\\ =\dfrac{x}{x\left(x-2\right)}-\dfrac{x^2+4x}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\left(ĐKXĐ:x\ne0;x\ne\pm2\right)\\ =\dfrac{x\left(x+2\right)-\left(x^2+4x\right)-2\left(x-2\right)}{x\left(x+2\right)\left(x-2\right)}\\ =\dfrac{x^2-x^2+2x-4x-2x+4}{x\left(x+2\right)\left(x-2\right)}\\ =\dfrac{-4x+4}{x\left(x+2\right)\left(x-2\right)}\)

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

2: \(C=A:B\)

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

\(C+1=\dfrac{2x+1+x^2}{x^2}=\dfrac{\left(x+1\right)^2}{x^2}>=0\)

=>C>=-1