a) \(2x^3+9x^2+7x-6=0\)
\(\Leftrightarrow2x^3+4x^2+5x^2+10x-3x-6=0\)
\(\Leftrightarrow2x^2\left(x+2\right)+5x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^2+5x-3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^2-x+6x-3\right)=0\) .
\(\Leftrightarrow\left(x+2\right)\left[2x\left(x+3\right)-\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)\left(x+3\right)=0\)
TH1: \(x+2=0\Rightarrow x=-2\)
TH2: \(2x-1=0\Rightarrow x=\frac{1}{2}\)
TH3: \(x+3=0\Rightarrow x=-3\)
b) \(\left(1-3x\right)^2=\left(5x+2\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}1-3x=5x+2\\1-3x=-5x-2\end{cases}}\)
TH1: \(1-3x=5x+2\)
\(\Leftrightarrow8x+2=1\)
\(\Leftrightarrow x=-\frac{1}{8}\)
TH2: \(1-3x=-5x-2\)
\(\Leftrightarrow-8x-2=1\)
\(\Leftrightarrow x=-\frac{3}{8}\)
c) \(x^2-6x+17=0\)
\(\Leftrightarrow x^2-6x+9+8=0\)
\(\Leftrightarrow\left(x-3\right)^2+8=0\)
\(\Leftrightarrow\left(x-3\right)^2=-8\)
Có: \(\left(x-3\right)^2\ge0\)
Vậy: Không tồn tại x