Question

GIÚP EM VÓI Ạ,EM CẦN GẤPPPPPP

Answer 1

ĐK : x \(\ge-1\)

Ta có : \(x^2-2x-1=\sqrt{\left(x^2+1\right)\left(x+1\right)}\)

<=> \(\left(x^2+1\right)-2\left(x+1\right)=\sqrt{\left(x^2+1\right)\left(x+1\right)}\)

Đặt \(\sqrt{x^2+1}=a;\sqrt{x+1}=b\)(\(a>0;b\ge0\))

Khi đó a² - 2b² = ab

<=> (a - 2b)(a + b) = 0

<=> a - 2b = 0

<=> a = 2b

<=> \(\sqrt{x^2+1}=2\sqrt{x+1}\)

<=> \(\left\{{}\begin{matrix}x^2+1=4x+4\\x\ge-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-4x-3=0\\x\ge-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=\sqrt{7}+2\\x=-\sqrt{7}+2\end{matrix}\right.\\x\ge-1\end{matrix}\right.\Leftrightarrow x=\sqrt{7}+2\)