\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)
\(\Leftrightarrow x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9=0\)
Đặt \(A=x\left(x+2\right)\left(x+4\right)\left(x+6\right)-9\)
\(A=x\left(x+6\right)\left(x+2\right)\left(x+4\right)-9\)
\(=\left(x^2+6x\right)\left(x^2+6x+8\right)-9\)(1)
Đặt \(a=x^2+6x\)
\(\Rightarrow\left(1\right)=a\left(a+8\right)-9=a^2+8a-9\)
\(=\left(a+4\right)^2-25=\left(a+4-5\right)\left(a+4+5\right)\)
\(=\left(a-1\right)\left(a+9\right)=\left(x^2+6x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x^2+6x-1\right)\left(x+3\right)^2\)
\(pt\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)
\(TH1:x^2+6x-1=0\)
\(\Leftrightarrow\left(x+3\right)^2-10=0\)
\(\Leftrightarrow\left(x+3\right)^2=10\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}}\Leftrightarrow x=\pm\sqrt{10}+3\)
\(TH2:\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
Vậy \(x\in\left\{\pm\sqrt{10}+3;-3\right\}\)
