Question

[Ngọc Hạnh]

giải pt: \(\left(x+3\right)\left(x+12\right)\left(x-4\right)\left(x-16\right)+20x^2=0\)

Answer 1

[Akai Haruma]

Lời giải:

Ta có:

\((x+3)(x+12)(x-4)(x-16)+20x^2=0\)

\(\Leftrightarrow [(x+3)(x-16)][(x+12)(x-4)]+20x^2=0\)

\(\Leftrightarrow (x^2-13x-48)(x^2+8x-48)+20x^2=0\)

Đặt \(x^2-12x-48=a\). PT trở thành:

\((a-x)(a+20x)+20x^2=0\)

\(\Leftrightarrow a^2+19ax-20x^2+20x^2=0\Leftrightarrow a^2+19ax=0\)

\(\Leftrightarrow a(a+19x)=0\)

\(\Leftrightarrow (x^2-12x-48)(x^2+7x-48)=0\)

\(\Leftrightarrow \left[\begin{matrix} x^2-12x-48=0\\ x^2+7x-48=0\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=6\pm 2\sqrt{21}\\ x=\frac{-7\pm \sqrt{241}}{2}\end{matrix}\right.\)

Vậy......