Lời giải:
\((x+1)(x+2)(x+3)(x+6)=8x^2\)
\(\Leftrightarrow [(x+1)(x+6)][(x+2)(x+3)]=8x^2\)
\(\Leftrightarrow (x^2+7x+6)(x^2+5x+6)=8x^2\)
Đặt \(x^2+5x+6=a\). Khi đó pt trở thành:
\((a+2x)a=8x^2\)
\(\Leftrightarrow a^2+2ax-8x^2=0\)
\(\Leftrightarrow a^2-2ax+4ax-8x^2=0\)
\(\Leftrightarrow a(a-2x)+4x(a-2x)=0\)
\(\Leftrightarrow (a-2x)(a+4x)=0\Rightarrow \left[\begin{matrix} a=2x\\ a=-4x\end{matrix}\right.\)
+) Nếu $a=2x$ thì:
\(x^2+5x+6=2x\Rightarrow x^2+3x+6=0\)
\(\Rightarrow (x+\frac{3}{2})^2+\frac{15}{4}=0\) (vô lý)
+) Nếu \(a=-4x\Rightarrow x^2+5x+6=-4x\)
\(\Rightarrow x^2+9x+6=0\)
\(\Rightarrow x=\frac{-9\pm \sqrt{57}}{2}\) (đều thỏa mãn)
Vậy..........
