Tìm x biết 2x^2+x-6=0

a) \(2x^2+x-6=0\Leftrightarrow2x^2+4x-3x-6=0\)

\(\Leftrightarrow2x\left(x+2\right)-3\left(x+2\right)=0\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2x=3\\x=-2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)

vậy \(x=\dfrac{3}{2};x=-2\)

b) \(-5x^2+17x-6=0\Leftrightarrow-5x^2+15x+2x-6=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}-5x+2=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}5x=2\\x=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\x=3\end{matrix}\right.\)

vậy \(x=\dfrac{2}{5};x=3\)

c) \(3x^2+22x-16=0\Leftrightarrow3x^2+24x-2x-16=0\)

\(\Leftrightarrow3x\left(x+8\right)-2\left(x+8\right)=0\Leftrightarrow\left(3x-2\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x=2\\x=-8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)

vậy \(x=\dfrac{2}{3};x=-8\)

d) \(2x^3+3x^2-8x+3=0\Leftrightarrow2x^3-3x^2+x+6x^2-9x+3=0\)

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

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

\(\left(x+3\right)\left(2x-1\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\2x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\2x=1\\x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\\x=1\end{matrix}\right.\) vậy \(x=-3;x=\dfrac{1}{2};x=1\)