\(\left(x+3\right)\left(x+1\right)\left(x+2\right)^2=72\)
\(\left(x^2+4x+3\right)\left(x^2+4x+4\right)=72\)
Đặt \(x^2+4x+3,5=t\)
=> \(\left(t-0,5\right)\left(t+0,5\right)=72\)
\(\Leftrightarrow t^2-0,25=72\)
\(\Leftrightarrow t^2=72,25\)
=> t=8,5 hoặc t=-8,5
=> \(x^2+4x+3,5=8,5\) hoặc \(x^2+4x+3,5=-8,5\)
\(x^2+4x-5=0\) \(x^2+4x+12=0\)
\(x^2+5x-x-5=0\) \(x^2+4x+4+8=0\)
\(\left(x+5\right)\left(x-1\right)=0\) \(\left(x+2\right)^2+8=0\) (vô lí -> vô nghiệm)
=> x=-5 hoặc x=1
Vậy....
\(\left(x+3\right)\left(x+1\right)\left(x+2\right)^2=72\)
⇔\(\left(x^2+4x+3\right)\left(x^2+4x+4\right)=72\)
Đặt t=\(x^2+4x+3,5\)
⇒(t+0,5)(t-0,5)=72
⇔\(t^2\)-0,25=72
⇔\(t^2\)=71,75
⇔\(\left(x^2+4x+3,5\right)^2\)=71,75
⇔\(x^2+4x+3,5\)=-8,47 hoặc 8,47
⇔\(\left[{}\begin{matrix}x^2+4x+3,5=-8,47\\x^2+4x+3,5=8,47\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}x^2+4x+11,97=0\\x^2+4x-4,97=0\end{matrix}\right.\)
⇔\(\left[{}\begin{matrix}\left(x+2\right)^2+7,9=0\left(loại\right)\\\left(x+2\right)^2=8,97\end{matrix}\right.\)
⇔x+2=-\(\sqrt{8,97}\)hoặc\(\sqrt{8,97}\)
⇔\(\left[{}\begin{matrix}x=-\sqrt{8,97}-2\\x=\sqrt{8,97}-2\end{matrix}\right.\)
