Question

(x+6)(3x+1)+x²-36=0

Answer 1

\(\left(x+6\right)\left(3x+1\right)+x^2-36=0\)

\(\Leftrightarrow\left(x+6\right)\left(3x+1\right)+\left(x-6\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(3x+1+x-6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(4x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x+6=0\\4x-5=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-6\\x=\frac{5}{4}\end{cases}}\)

Vậy \(x=-6;\frac{5}{4}\)