Bài 35: 

b) ĐKXĐ: \(x\notin\left\{5;2\right\}\)

Ta có: \(\dfrac{x+2}{x-5}+3=\dfrac{6}{2-x}\)

\(\Leftrightarrow\dfrac{x+2}{x-5}+3-\dfrac{6}{2-x}=0\)

\(\Leftrightarrow\dfrac{x+2}{x-5}+3+\dfrac{6}{x-2}=0\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\dfrac{3\left(x-5\right)\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\dfrac{6\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=0\)

Suy ra: \(x^2-4+3\left(x^2-7x+10\right)+6x-30=0\)

\(\Leftrightarrow x^2-4+3x^2-21x+30+6x-30=0\)

\(\Leftrightarrow4x^2-15x-4=0\)

\(\Leftrightarrow4x^2-16x+x-4=0\)

\(\Leftrightarrow4x\left(x-4\right)+\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\4x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\4x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{4;-\dfrac{1}{4}\right\}\)

Bài 36: 

a) Ta có: \(\left(3x^2-5x+1\right)\left(x^2-4\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(3x^2-5x+1\right)=0\)

mà \(3x^2-5x+1>0\forall x\)

nên (x-2)(x+2)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy: S={2;-2}

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