Question

\(\text{(3x+y)^3-3(3x+y)^2+3(3x+y)-1=-27 }\)

Answer 1

Đặt \(t=3x+y\)

pt \(\Leftrightarrow t^3-3t^2+3t-1=-27\)

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

\(\Leftrightarrow\left(t-1+3\right)\left(\left(t-1\right)^2-3\left(t-1\right)+9\right)=0\)

\(\Leftrightarrow\left(t+2\right)\left(t^2-5t+13\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t+2=0\\t^2-5t+13=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-2\\t^2-5t+13=0\left(vl\right)\end{matrix}\right.\)

\(\Leftrightarrow3x+y=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-2-3x\\x\in R\end{matrix}\right.\)