View attachment 105487
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2,
Đặt [tex]\sqrt{x+1}= a \left ( a\geq 1 \right )[/tex]
pt [tex] \Leftrightarrow a^{2}= \left ( 2a^{2}-1 \right )\sqrt{a+2}[/tex]
[tex]\Leftrightarrow \left ( 2a^{2}-1 \right )\left ( \sqrt{a+2} -2a\right )+4a^{3}-a^{2}-2a= 0[/tex]
[tex] \left ( 4a^{2}+a+2 \right )\frac{-1-a\sqrt{a+2}}{\sqrt{a+2}+2a}= 0[/tex]
Ta có:[tex] \frac{-1-a\sqrt{a+2}}{\sqrt{a+2}+2a}= 0\Leftrightarrow -1-a\sqrt{a+2}= 0\Leftrightarrow a\sqrt{a+2}= -1\left ( vn \right )[/tex]
Suy ra :[tex] -4a^{2}+a+2= 0\Leftrightarrow a= \frac{1+\sqrt{33}}{8}\left ( a\geq 0 \right )[/tex]
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