\(\Leftrightarrow x^2+3-\left(6x+1\right)\sqrt{x^2+3}+9x^2+3x-2=0\)
Đặt \(\sqrt{x^2+3}=t\)
\(\Rightarrow t^2-\left(6x+1\right)t+9x^2+3x-2=0\)
\(\Delta=\left(6x+1\right)^2-4\left(9x^2+3x-2\right)=9\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{6x+1+3}{2}=3x+2\\t=\frac{6x+1-3}{2}=3x-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+2}=3x+2\left(x\ge-\frac{2}{3}\right)\\\sqrt{x^2+2}=3x-1\left(x\ge\frac{1}{3}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2=\left(3x+2\right)^2\\x^2+2=\left(3x-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow...\)
Đúng 0
Bình luận (0)