TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;-1\right\}\end{matrix}\right.\)

1:

\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)

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

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

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

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

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

\(=-\dfrac{1}{x+2}\left(x\ne2;x\ne-2\right)\)

\(\dfrac{-3}{x+2}-\dfrac{2}{x-2}+\dfrac{4x}{x^2-4}\left(x\ne\pm2\right)\)

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

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

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

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

TXĐ: \(\left\{{}\begin{matrix}x\in R\\x\notin\left\{0;2;-2\right\}\end{matrix}\right.\)

Ta có: \(\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)

\(=\dfrac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-4+10-x^2}{x+2}\)

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

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{6}\)

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

a, đề này chắc sai ở đoạn \(\dfrac{2x}{x^2-3}\) sửa thành \(\dfrac{2x}{x-3}\)

\(=>đk:x\ne1,x\ne3\)

\(=>A=\dfrac{2x}{x-3}+\dfrac{2x}{x^2-4x+3}+\dfrac{x}{x-1}\)

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

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

b, \(A=\dfrac{3x}{x-3}=3+\dfrac{9}{x-3}\)

A nguyên <=>\(x-3\inƯ\left(9\right)=\left\{\pm1;\pm3;\pm9\right\}\)

\(=>x\in\left\{4;2;6;0;12;-6\right\}\left(TM\right)\)

a: ĐKXĐ: x<>2; x<>0

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

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

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

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

c: M>=-3

=>(x+1+6x)/2x>=0

=>(7x+1)/x>=0

=>x>0 hoặc x<=-1/7