\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+15=0\)
\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+15=0\)
Đặt \(x^2+10+16=t\) , ta được:
\(t\left(t+8\right)+15=0\)
\(\Leftrightarrow t^2+8t+15=0\)
\(\Leftrightarrow t^2+8t+16=1\)
\(\Leftrightarrow\left(t+4\right)^2=1\Rightarrow\left[{}\begin{matrix}t+4=1\\t+4=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}t=-3\\t=-5\end{matrix}\right.\)
với t = -3 ta có:
\(x^2+10x+16=-3\)
\(\Leftrightarrow\left(x^2+10x+25\right)=6\)
\(\Leftrightarrow\left(x+5\right)^2=6\)
\(\Rightarrow\left[{}\begin{matrix}x+5=\sqrt{6}\\x+5=-\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}-5\\x=-\sqrt{6}-5\end{matrix}\right.\)
Với t =-5 ta có:
\(x^2+10x+16=-5\)
\(\Leftrightarrow x^2+10x+25=4\)
\(\Leftrightarrow\left(x+5\right)^2=4\)
\(\Rightarrow\left[{}\begin{matrix}x+5=2\\x+5=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-7\end{matrix}\right.\)
