Bài 1 :
\(a,2x-4=8\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\)
Vậy \(S=\left\{6\right\}\)
\(b,\dfrac{5x-3}{2}=\dfrac{5-2x}{3}\)
\(\Leftrightarrow3\left(5x-3\right)=2\left(5-2x\right)\)
\(\Leftrightarrow15x-9=10-4x\)
\(\Leftrightarrow15x+4x=10+9\)
\(\Leftrightarrow19x=19\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
\(c,\left(x^2-4\right)+\left(x-2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\-2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
\(d,2x-\dfrac{2x^2}{x+3}=\dfrac{4x}{x+3}+\dfrac{2}{7}\left(ĐKXĐ:x\ne-3\right)\)
\(\Leftrightarrow14x\left(x+3\right)-14x^2=28x+2\left(x+3\right)\)
\(\Leftrightarrow14x^2+42x-14x^2=28x+2x+6\)
\(\Leftrightarrow14x^2-14x^2+42x-28x-2x-6=0\)
\(\Leftrightarrow12x-6=0\)
\(\Leftrightarrow12x=6\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(nhận\right)\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
