Question

Tìm x<0, biết: \(\left(8x^2+3\right)\left(8x^2-3\right)-\left(8x^2-1^2\right)=54\)

Answer 1

Đặt 8x^2=t => t>=0

\(t^2-9-\left(t-1\right)=54\Leftrightarrow t^2-t+1-9=54\)

\(t^2-t+\frac{1}{4}=54+8+\frac{1}{4}=\frac{249}{4}\) lẻ thế nhỉ

\(\left(t-\frac{1}{2}\right)^2=\frac{249}{4}\Rightarrow\left[\begin{matrix}t=\frac{1-\sqrt{249}}{2}< 0\left(loai\right)\\t=\frac{1+\sqrt{249}}{2}\end{matrix}\right.\)

\(\left\{\begin{matrix}x< 0\\8x^2=\frac{1+\sqrt{249}}{2}\end{matrix}\right.\Rightarrow x=\frac{-\sqrt{1+\sqrt{249}}}{16}\)