1)\(\dfrac{1}{2-x}=\dfrac{1}{2x^2+1}\) (đk: \(x\ne2\))
\(\Leftrightarrow2-x=2x^2+1\)
\(\Leftrightarrow2x^2+x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)(Thỏa)
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4) \(\left|3x+1\right|+x=2\)
\(\Leftrightarrow\left|3x+1\right|=2-x\)
TH1: \(3x+1\ge0\Leftrightarrow x\ge-\dfrac{1}{3}\)
PT \(\Leftrightarrow3x+1=2-x\)
\(\Leftrightarrow x=\dfrac{1}{4}\) (tm)
TH2:\(3x+1< 0\Leftrightarrow x< -\dfrac{1}{3}\)
PT \(\Leftrightarrow-3x-1=2-x\)
\(\Leftrightarrow x=-\dfrac{3}{2}\) (tm)
Vậy pt có tập nghiệm \(S=\left\{-\dfrac{3}{2};\dfrac{1}{4}\right\}\)
6) \(\left|x^2-5x+4\right|=\left|x^2+x\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x+4=x^2+x\\x^2-5x+4=-x^2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\2x^2-4x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\2\left(x-1\right)^2+2=0\left(vn\right)\end{matrix}\right.\)
Vậy...
