9 tháng 2 2020 lúc 23:30
ĐKXĐ:...
Đặt \(\sqrt{x+1}=a\ge0\)
\(\Leftrightarrow a^2=\left(2a^2-1\right)\sqrt{a+2}\)
\(\Leftrightarrow\left(2a^2-1\right)\left(2a-\sqrt{a+2}\right)=-a^2+2a\left(2a^2-1\right)\)
\(\Leftrightarrow\frac{\left(2a^2-1\right)\left(4a^2-a-2\right)}{2a+\sqrt{a+2}}=a\left(4a^2-a-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4a^2-a-2=0\\\frac{2a^2-1}{2a+\sqrt{a+2}}=a\left(1\right)\end{matrix}\right.\)
Xét (1), ta có:
\(\frac{2a^2-1}{2a+\sqrt{a+2}}=a-\frac{1+a\sqrt{a+2}}{2a+\sqrt{a+2}}< a\)
\(\Rightarrow\left(1\right)\) vô nghiệm
