$ĐKXĐ : x \neq 2, x \neq -2$

Ta có : $1+\dfrac{2}{x-2} = \dfrac{2x^2}{x^2-4}$

$\to \dfrac{x^2-4+2.(x+2)}{(x-2).(x+2)} = \dfrac{2x^2}{(x-2).(x+2)}$

$\to x^2-4+2.(x+2)  = 2x^2$

$\to x^2 -2x - 8 = 0 $

$\to (x-4).(x+2) = 0 $

$\to x = 4$ ( Do $x \neq -2, 2$ )

Vậy \(S=\left\{4\right\}\)

\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)

\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)

\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)

đúng hay không thì thay vào là được mà =))

mình bấm máy ra -1 b

\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)

a: (x-3)(x-1)-x(x-2)=0

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

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

=>-2x=-3

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

b: \(\left(x+2y\right)^2-\left(2x-y\right)^2\)

\(=\left(x+2y+2x-y\right)\left(x+2y-2x+y\right)\)

\(=\left(3x+y\right)\left(-x+3y\right)\)

\(a)\)

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

\(\Leftrightarrow x-x^2+1=3x+1\)

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

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(b)\)

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

\(\Leftrightarrow x^2+4x+4-2x-3=x^2+10\)

\(\Leftrightarrow x^2+2x+1=x^2+10\)

\(\Leftrightarrow2x-9=0\)

\(\Leftrightarrow2x=9\)

\(\Leftrightarrow x=\frac{2}{9}\)

\(a.\left(2x-1\right)^2-\left(4x-3\right)\left(x+5\right)=0\)  \(\Leftrightarrow4x^2-4x+1-\left(4x^2+17x-15\right)=0\)

\(\Leftrightarrow-21x+16=0\Leftrightarrow x=\dfrac{16}{21}\) . Vậy ... 

b.\(x\left(x-1\right)=3\left(x-1\right)\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)  \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) . Vậy ...

c.\(\left(x-1\right)\left(3x-7\right)=\left(x-1\right)\left(x+3\right)\Leftrightarrow\left(x-1\right)\left(3x-7-x-3\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\) . Vậy ... 

d.\(\left(x-3\right)^2+2x-6=0\Leftrightarrow\left(x-3\right)\left(x-3+2\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) . Vậy ... 

cảm ơn bạn nhiều ạ