a) Ta có: (x-5)(x+5)=2x-5
\(\Leftrightarrow x^2-25-2x+5=0\)
\(\Leftrightarrow x^2-2x-20=0\)
\(\Leftrightarrow x^2-2x+1=21\)
\(\Leftrightarrow\left(x-1\right)^2=21\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=\sqrt{21}\\x-1=-\sqrt{21}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{21}+1\\x=-\sqrt{21}+1\end{matrix}\right.\)
Vậy: \(S=\left\{\sqrt{21}+1;-\sqrt{21}+1\right\}\)
b) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x}{x-1}-\dfrac{2x}{x^2-1}=0\)
\(\Leftrightarrow\dfrac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
Suy ra: \(x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=1\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}
c)ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{2x}{x+1}-\dfrac{4}{x^2-1}=\dfrac{2x-5}{x-1}\)
\(\Leftrightarrow\dfrac{2x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(2x-5\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(2x^2-2x-4=2x^2+2x-5x-5\)
\(\Leftrightarrow-2x-4-2x+5x+5=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1(loại)
Vậy: \(S=\varnothing\)
d)ĐKXĐ: \(x\notin\left\{-2;-3\right\}\)
Ta có: \(\dfrac{5\left(x-2\right)}{x+2}-\dfrac{2\left(x-3\right)}{x+3}=3\)
\(\Leftrightarrow\dfrac{5\left(x-2\right)\left(x+3\right)}{\left(x+2\right)\left(x+3\right)}-\dfrac{2\left(x-3\right)\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}=\dfrac{3\left(x+3\right)\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}\)
Suy ra: \(5\left(x^2+x-6\right)-2\left(x^2-x-6\right)=3\left(x^2+5x+6\right)\)
\(\Leftrightarrow5x^2+5x-30-2x^2+2x+12=3x^2+15x+18\)
\(\Leftrightarrow3x^2+7x-18-3x^2-15x-18=0\)
\(\Leftrightarrow-8x-36=0\)
\(\Leftrightarrow-8x=36\)
hay \(x=-\dfrac{9}{2}\)(thỏa ĐK)
Vậy: \(S=\left\{-\dfrac{9}{2}\right\}\)
e) Ta có: \(\left|2x-10\right|=-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-10=-3x\left(x\ge5\right)\\-2x+10=-3x\left(x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3x-10=0\\-2x+3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=10\\x=-10\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(loại\right)\\x=-10\left(nhận\right)\end{matrix}\right.\)
Vậy: S={-10}
