\(ĐK:x\ne3;x\ne2\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(\dfrac{x+1}{x-2}+1+\dfrac{x^2}{x-2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\\dfrac{x^2+x+2}{x-2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x^2+x+2=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

ĐKXĐ: \(x\notin\left\{2;-1;\dfrac{-3\pm\sqrt{17}}{2}\right\}\)

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

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

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

=>\(4x^3-8x=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)

=>\(4x\left(x^2-2\right)=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)

=>\(\left(x^2-2\right)^2-2x\left(x^2-2\right)-3x^2=0\)

=>\(\left(x^2-2\right)^2-3x\left(x^2-2\right)+x\left(x^2-2\right)-3x^2=0\)

=>\(\left(x^2-2\right)\left(x^2-2-3x\right)+x\left(x^2-2-3x\right)=0\)

=>\(\left(x^2+x-2\right)\left(x^2-3x-2\right)=0\)

=>\(\left(x+2\right)\left(x-1\right)\left(x^2-3x-2\right)=0\)

=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=1\left(nhận\right)\\x=\dfrac{3\pm\sqrt{17}}{2}\left(nhận\right)\end{matrix}\right.\)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

\(\Leftrightarrow\dfrac{3+2x^2-2x-x+3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=0\)

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

\(\Leftrightarrow x\left(4x-3\right)-\left(x-2\right)\left(3x+2\right)=x^2-5\)

\(\Leftrightarrow4x^2-3x-3x^2-2x+6x+4=x^2-5\)

\(\Leftrightarrow x^2+x+4=x^2-5\)

=>x+4=-5

hay x=-9(nhận)

Sửa đề: \(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)

ĐKXĐ: \(x\notin\left\{0;-2\right\}\)

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

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

=>\(x\left(2x-1\right)+3x+2=\left(x+1\right)\left(x+2\right)\)

=>\(2x^2-x+3x+2=x^2+3x+2\)

=>\(2x^2+2x-x^2-3x=0\)

=>\(x^2-x=0\)

=>x(x-1)=0

=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

\(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\\ \Leftrightarrow\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=-\dfrac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}\\ đkxđ:\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\\ \Leftrightarrow\dfrac{\left(x^2-x\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\\ \Leftrightarrow\dfrac{0}{\left(x-3\right)\left(x+3\right)}=0\\ \Rightarrow0=0\left(luon.dung\right)\)

ĐKXĐ: \(x\ne\pm2\)

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

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

\(\Rightarrow x^2-2x=4x^2-x-4-3x^2-5x+2\)

\(\Leftrightarrow4x=-2\)

\(\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

Vậy...

1) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)

\(\Leftrightarrow\dfrac{\left(1-6x\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{\left(9x+4\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{3x^2-2x+1}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=3x^2-2x+1\)

\(\Leftrightarrow x+2-6x^2-12x+9x^2-18x+4x-8-3x^2+2x-1=0\)

\(\Leftrightarrow-23x-7=0\)

\(\Leftrightarrow-23x=7\)

\(\Leftrightarrow x=-\dfrac{7}{23}\)(nhận)

Vậy: \(S=\left\{-\dfrac{7}{23}\right\}\)

2) ĐKXĐ: \(x\notin\left\{\dfrac{2}{3};-\dfrac{2}{3}\right\}\)

Ta có: \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2-3x}=\dfrac{9x^2}{9x^2-4}\)

\(\Leftrightarrow\dfrac{3x+2}{3x-2}+\dfrac{6}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{3x+8}{3x-2}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

\(\Leftrightarrow\dfrac{\left(3x+8\right)\left(3x+2\right)}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{9x^2}{\left(3x-2\right)\left(3x+2\right)}\)

Suy ra: \(9x^2+6x+24x+16=9x^2\)

\(\Leftrightarrow30x+16=0\)

\(\Leftrightarrow30x=-16\)

hay \(x=-\dfrac{8}{15}\)(nhận)

Vậy: \(S=\left\{-\dfrac{8}{15}\right\}\)

ĐK: ` x \ne \pm 3`

`(x^2-x)/(x+3)-(x^2)/(x-3)=(7x^2-3x)/(9-x^2)`

`<=> (x^2-x)(x-3)-x^2 (x+3) = -(7x^2-3x)`

`<=> −7x^2+3x=-7x^2+3x`

`<=> 0x=0 forall x`

Vậy `S=RR \\ {+-3}`.