p) \(\left|x-4\right|+\left|x+1\right|=9\)
-Có: \(\left[{}\begin{matrix}\left|x-4\right|=x-4\Leftrightarrow x\ge4\\\left|x-4\right|=4-x\Leftrightarrow x< 4\end{matrix}\right.\) ; \(\left[{}\begin{matrix}\left|x+1\right|=x+1\Leftrightarrow x\ge-1\\\left|x+1\right|=-x-1\Leftrightarrow x< -1\end{matrix}\right.\)
-Xét \(x< -1:\)
PT \(\Leftrightarrow4-x+\left(-x-1\right)=9\)
\(\Leftrightarrow4-x-x-1=9\)
\(\Leftrightarrow-2x+3=9\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\left(nhận\right)\)
-Xét \(-1\le x< 4:\)
PT \(\Leftrightarrow4-x+x+1=9\)
\(\Leftrightarrow5=9\) (vô lý)
-Xét \(x\ge4:\)
PT \(\Leftrightarrow x-4+x+1=9\)
\(\Leftrightarrow2x-3=9\)
\(\Leftrightarrow2x=12\)
\(\Leftrightarrow x=6\left(nhận\right)\)
-Vậy \(S=\left\{-3;6\right\}\)
h) \(2x^3-2x^2+x=1\)
\(\Leftrightarrow2x^3-2x^2+x-1=0\)
\(\Leftrightarrow2x^2\left(x-1\right)+x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2+1\right)=0\)
\(\Leftrightarrow x-1=0\) hay \(2x^2+1=0\) (pt vô nghiệm do \(2x^2+1\ge1>0\forall x\))
\(\Leftrightarrow x=1\)
-Vậy \(S=\left\{1\right\}\)
