a) \(3x^2+11x+6=10\)
\(\Leftrightarrow3x^2+11x-4=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-4\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{3};-4\right\}\)
b) \(5x^2-13x+6=0\)
\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=2\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{5};2\right\}\)
`a)`\(3x^2+11x+6=10\)
\(\Rightarrow3x^2+11x-4=0\)
\(\Rightarrow3x^2-x+12x-4=0\)
\(\Rightarrow3x\left(3x-1\right)+4\left(3x-1\right)=0\)
\(\Rightarrow\left(x+4\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{1}{3}\end{matrix}\right.\)
`b)`\(5x^2-13x+6=0\)
\(\Rightarrow5x^2-10x-3x+6=0\)
\(\Rightarrow5x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(5x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)
