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$ĐKXĐ : x \neq 2, x \neq -2$
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 \(S=\left\{4\right\}\)
\(a,3x-2\left(x-3\right)=0\\ \Leftrightarrow3x-2x+6=0\\ \Leftrightarrow x=-6\\ b,\left(x+1\right)\left(2x-3\right)=\left(2x-1\right)\left(x+5\right)\\ \Leftrightarrow2x^2+2x-3x-3=2x^2-x+10x-5\\ \Leftrightarrow2x^2-x-3=2x^2+9x-5\\ \Leftrightarrow10x-2=0\\ \Leftrightarrow x=\dfrac{1}{5}\\ c,ĐKXĐ:x\ne\pm1\\ \dfrac{2x}{x-1}-\dfrac{x}{x+1}=1\\ \Leftrightarrow\dfrac{2x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\\ \Leftrightarrow\dfrac{2x^2+2x-x^2+x-x^2+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow3x+1=0\\ \Leftrightarrow x=-\dfrac{1}{3}\left(tm\right)\)
\(d,\left(2x+3\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x-5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{5}{3}\end{matrix}\right.\\ e,ĐKXĐ:x\ne\pm2\\ \dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \Leftrightarrow\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-22}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\dfrac{x^2-4x+4-3x-6-2x+22}{\left(x-2\right)\left(x+2\right)}=0\\ \Rightarrow x^2-9x+20=0\\ \Leftrightarrow\left(x^2-5x\right)-\left(4x-20\right)=0\\ \Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)\\ \Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=5\left(tm\right)\end{matrix}\right.\)
\(a,\Leftrightarrow x^2-2x-x^2+1=0\\ \Leftrightarrow-2x+1=0\Leftrightarrow x=\dfrac{1}{2}\\ b,\Leftrightarrow\left(2x-1-x-4\right)\left(2x-1+x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(3x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
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 ...
\(\dfrac{2}{x}=\dfrac{x}{x+1}\left(ĐKXĐ:x\ne0;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{2\left(x+1\right)}{x\left(x+1\right)}=\dfrac{x^2}{x\left(x+1\right)}\)
\(\Rightarrow x^2=2x+2\)
\(\Leftrightarrow x^2-2x-2=0\)
\(\Leftrightarrow x^2-2x+1-3=0\)
\(\Leftrightarrow\left(x-1\right)^2-3=0\)
\(\Leftrightarrow\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{3}=0\\x-1+\sqrt{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+\sqrt{3}\left(nhận\right)\\x=1-\sqrt{3}\left(nhận\right)\end{matrix}\right.\)
-Vậy \(S=\left\{1+\sqrt{3};1-\sqrt{3}\right\}\)
\(\dfrac{2}{x}=\dfrac{x}{x+1}\left(x\ne0;-1\right)\) \(\Leftrightarrow2x+2=x^2\Leftrightarrow x^2-2x-2=0\) \(\Leftrightarrow\left(x-1\right)^2=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{matrix}\right.\) . Vậy ...