a/Thay x=1 đc \(k^2+5k+6=0\Leftrightarrow k^2+2k+3k+6=0\)
\(\Leftrightarrow k\left(k+2\right)+3\left(k+2\right)=0\Leftrightarrow\left(k+3\right)\left(k+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}k=-2\\k=-3\end{matrix}\right.\)
b/Thay k=-2 đc \(4x^2-10x+6=0\Leftrightarrow2x^2-5x+3=0\)
\(\Leftrightarrow2x^2-2x-3x+3=0\Leftrightarrow2x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)
Thay k=-3 đc \(9x^2-15x+6=0\Leftrightarrow3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=1\end{matrix}\right.\)
