Đáp án: x=0 hoặc x=1
Giải thích các bước giải:
$\begin{array}{l}
Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
x + 1 \ge 0
\end{array} \right. \Rightarrow x \ge 0\\
\left( {\sqrt x - \sqrt {x + 1} } \right)\left( {\sqrt {{x^2} + x} + 1} \right) = - 1\\
\Rightarrow \left( {\sqrt x + \sqrt {x + 1} } \right)\left( {\sqrt x - \sqrt {x + 1} } \right).\left( {\sqrt {{x^2} + x} + 1} \right) = - \left( {\sqrt x + \sqrt {x + 1} } \right)\\
\Rightarrow \left( {x - x - 1} \right).\left( {\sqrt {{x^2} + x} + 1} \right) = - \left( {\sqrt x + \sqrt {x + 1} } \right)\\
\Rightarrow \sqrt {{x^2} + x} + 1 = \sqrt x + \sqrt {x + 1} \\
\Rightarrow \sqrt {x\left( {x + 1} \right)} + 1 - \sqrt x - \sqrt {x + 1} = 0\\
\Rightarrow \sqrt x \left( {\sqrt {x + 1} - 1} \right) - \left( {\sqrt {x + 1} - 1} \right) = 0\\
\Rightarrow \left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {x + 1} = 1\\
\sqrt x = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x + 1 = 1\\
x = 1
\end{array} \right.\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\left( {tmdk} \right)
\end{array}$
