Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
6,\\
2\sqrt {x + 2 + 2\sqrt {x + 1} } - \sqrt {x + 1} = 4\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - 1} \right)\\
\Leftrightarrow 2.\sqrt {\left( {x + 1} \right) + 2\sqrt {x + 1} + 1} - \sqrt {x + 1} = 4\\
\Leftrightarrow 2.\sqrt {{{\left( {\sqrt {x + 1} + 1} \right)}^2}} - \sqrt {x + 1} = 4\\
\Leftrightarrow 2.\left( {\sqrt {x + 1} + 1} \right) - \sqrt {x + 1} = 4\\
\Leftrightarrow \sqrt {x + 1} + 2 = 4\\
\Leftrightarrow \sqrt {x + 1} = 2\\
\Leftrightarrow x = 3\\
7,\\
\sqrt x + \sqrt {x + 1} = 1 + \sqrt {x\left( {x + 1} \right)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 0} \right)\\
\Leftrightarrow \sqrt x .\sqrt {x + 1} + 1 - \sqrt x - \sqrt {x + 1} = 0\\
\Leftrightarrow \left( {\sqrt x .\sqrt {x + 1} - \sqrt x } \right) - \left( {\sqrt {x + 1} - 1} \right) = 0\\
\Leftrightarrow \sqrt x .\left( {\sqrt {x + 1} - 1} \right) - \left( {\sqrt {x + 1} - 1} \right) = 0\\
\Leftrightarrow \left( {\sqrt {x + 1} - 1} \right)\left( {\sqrt x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x + 1} = 1\\
\sqrt x = 1
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
8,\\
x + 2\sqrt {7 - x} = 2\sqrt {x - 1} + \sqrt { - {x^2} + 8x - 7} + 1\,\,\,\,\,\,\,\,\,\,\,\left( {1 \le x \le 7} \right)\\
\Leftrightarrow \left( {x - 1} \right) + 2.\left( {\sqrt {7 - x} - \sqrt {x - 1} } \right) - \sqrt { - {x^2} + 8x - 7} = 0\\
\Leftrightarrow \left( {x - 1} \right) + 2.\left( {\sqrt {7 - x} - \sqrt {x - 1} } \right) - \sqrt {\left( {x - 1} \right)\left( {7 - x} \right)} = 0\\
\Leftrightarrow \sqrt {x - 1} \left( {\sqrt {x - 1} - \sqrt {7 - x} } \right) - 2.\left( {\sqrt {x - 1} - \sqrt {7 - x} } \right) = 0\\
\Leftrightarrow \left( {\sqrt {x - 1} - \sqrt {7 - x} } \right)\left( {\sqrt {x - 1} - 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt {x - 1} = \sqrt {7 - x} \\
\sqrt {x - 1} = 2
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x - 1 = 7 - x\\
x - 1 = 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 4\\
x = 5
\end{array} \right.
\end{array}\)
