Đặt \(\sqrt{x^2+1}=a\ge1\) pt trở thành:
\(\left(4x-1\right)a=2a^2+2x-1\)
\(\Leftrightarrow2a^2+a-1-4ax+2x=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a+1\right)-2x\left(2a-1\right)=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a+1-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\frac{1}{2}< 1\left(l\right)\\a=2x-1\end{matrix}\right.\) \(\Rightarrow\sqrt{x^2+1}=2x-1\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\x^2+1=\left(2x-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\3x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0< \frac{1}{2}\left(l\right)\\x=\frac{4}{3}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
