Lời giải:
PT \(\Leftrightarrow [(x-5)(x-8)][(x-4)(x-10)]=72x^2\)
\(\Leftrightarrow (x^2-13x+40)(x^2-14x+40)=72x^2\)
Đặt \(x^2-13x+40=a\) thì pt trở thành:
\(a(a-x)=72x^2\)
\(\Leftrightarrow a^2-ax-72x^2=0\)
\(\Leftrightarrow a^2-9ax+8ax-72x^2=0\)
\(\Leftrightarrow a(a-9x)+8x(a-9x)=0\)
\(\Leftrightarrow (a-9x)(a+8x)=0\)
Nếu $a-9x=0$
\(\Leftrightarrow x^2-13x+40-9x=0\)
\(\Leftrightarrow x^2-22x+40=0\)
\(\Leftrightarrow (x-2)(x-20)=0\Rightarrow \left[\begin{matrix} x=2\\ x=20\end{matrix}\right.\)
Nếu $a+8x=0$
\(\Leftrightarrow x^2-13x+40+8x=0\)
\(\Leftrightarrow x^2-5x+40=0\Leftrightarrow (x-\frac{5}{2})^2=-\frac{135}{4}\) (vô lý)
Vậy........
