Question

Giải phương trình: \((x-5)(x-4)(x-8)(x-10)=72x^2\)

Answer 1

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........