\(2x^3-5x^2+5x-2=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x^2+x+1\right)-5x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2+2x+2-5x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-3x+2\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-1=0\\x^2-3x+2=0\left(loai\right)\end{matrix}\right.\)\(\Leftrightarrow x=1\)
\(2x^3-5x^2+5x-2=0\)
\(\Leftrightarrow2x^3-2x^2-3x^2+3x+2x-2=0\)
\(\Leftrightarrow2x^2\left(x-1\right)-3x\left(x-1\right)+2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^2-3x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^2-2x+1\right)+\left(x^2-x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]=0\)
\(\Leftrightarrow\left[\begin{matrix}x-1=0\\\left(x-1\right)^2+\left(x-\frac{1}{2}\right)^2+\frac{3}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\Ta luôn có: \left(x-1\right)^2\ge0; \left(x-\frac{1}{2}\right)^2\ge0; \frac{3}{4}>0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=1\\Pt vô nghiệm\end{matrix}\right.\)
Vậy tập nghiệm của pt là S={1}
