a) Ta có: 3(x+1)-(4x-3)=5
\(\Leftrightarrow3x+3-4x+3-5=0\)
\(\Leftrightarrow-x+1=0\)
hay x=1
Vậy: S={1}
b) Ta có: \(\left(2x+1\right)\left(x-1\right)=x^2-2x+1\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=\left(x-1\right)^2\)
\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1-x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
Vậy: S={1;-2}
c) Ta có: |-5x|=3x-16(*)
Trường hợp 1: \(-5x\ge0\Leftrightarrow x\le0\)
(*)\(\Leftrightarrow-5x=3x-16\)
\(\Leftrightarrow-5x-3x=-16\)
\(\Leftrightarrow-8x=-16\)
hay x=2(loại)
Trường hợp 2: \(-5x< 0\Leftrightarrow x>0\)
(*)\(\Leftrightarrow5x=3x-16\)
\(\Leftrightarrow5x-3x=-16\)
\(\Leftrightarrow2x=-16\)
hay x=-8(loại)
Vậy: \(S=\varnothing\)
d) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\frac{x-2}{x-2}-\frac{2}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{x-4}{x-2}=\frac{5x+2}{4-x^2}\)
\(\Leftrightarrow\frac{\left(x-4\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{5x+2}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{x^2-2x-8+5x+2}{\left(x-2\right)\left(x+2\right)}=0\)
Suy ra: \(x^2+3x-6=0\)
\(\Leftrightarrow x^2+2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{33}{4}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{33}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{2}=\frac{\sqrt{33}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{33}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{33}-3}{2}\left(tm\right)\\x=\frac{-\sqrt{33}-3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy: \(S=\left\{\frac{\sqrt{33}-3}{2};\frac{-\sqrt{33}-3}{2}\right\}\)
