a: \(6\left(x-1\right)-5=3x\)
=>\(6x-6-5-3x=0\)
=>3x-11=0
=>3x=11
=>\(x=\dfrac{11}{3}\)
b: \(\left(2x+5\right)^2-\left(x-2\right)^2=0\)
=>\(\left(2x+5+x-2\right)\left(2x+5-x+2\right)=0\)
=>(3x+3)(x+7)=0
=>3(x+1)(x+7)=0
=>(x+1)(x+7)=0
=>\(\left[{}\begin{matrix}x+1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-7\end{matrix}\right.\)
c: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)
=>\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{-5x+2}{\left(x-2\right)\cdot\left(x+2\right)}\)
=>\(\dfrac{\left(x-1\right)\left(x-2\right)-x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{-5x+2}{\left(x-2\right)\left(x+2\right)}\)
=>\(x^2-3x+2-x^2-2x=-5x+2\)
=>-5x+2=-5x+2
=>0x=0(luôn đúng)
Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}