Question

b)(x^2+2x+3)^2 - 9(x^2+2x+3) + 18 = 0

Answer 1

\(\left(x^2+2x+3\right)^2-9.\left(x^2+2x+3\right)+18=0\)

\(\Leftrightarrow\left(x^2+2x+3\right)^2-3\left(x^2+2x+3\right)-6\left(x^2+2x+3\right)+18=0\)

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

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

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

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