Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
\sqrt x + \sqrt {1 - x} = 1\,\,\,\,\,\,\,\,\,\,\,\,\left( {0 \le x \le 1} \right)\\
\Leftrightarrow {\left( {\sqrt x + \sqrt {1 - x} } \right)^2} = 1\\
\Leftrightarrow x + 2\sqrt {x\left( {1 - x} \right)} + \left( {1 - x} \right) = 1\\
\Leftrightarrow 1 + 2\sqrt {x\left( {1 - x} \right)} = 1\\
\Leftrightarrow \sqrt {x\left( {1 - x} \right)} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
1 - x = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
b,\\
\sqrt {3 - x} + \sqrt {x - 5} = 10\\
DK:\,\,\,\,\left\{ \begin{array}{l}
3 - x \ge 0\\
x - 5 \ge 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le 3\\
x \ge 5
\end{array} \right.\\
\Rightarrow ptvn\\
c,\\
\sqrt {x + \sqrt {2x - 1} } + \sqrt {x - \sqrt {2x - 1} } = \sqrt 2 \,\,\,\,\,\,\,\,\left( {x \ge \frac{1}{2}} \right)\\
\Leftrightarrow \sqrt {\frac{1}{2}.\left( {2x + 2\sqrt {2x - 1} } \right)} + \sqrt {\frac{1}{2}\left( {2x - 2\sqrt {2x - 1} } \right)} = \sqrt 2 \\
\Leftrightarrow \sqrt {\frac{1}{2}.\left[ {\left( {2x - 1} \right) + 2\sqrt {2x - 1} + 1} \right]} + \sqrt {\frac{1}{2}\left[ {\left( {2x - 1} \right) - 2\sqrt {2x - 1} + 1} \right]} = \sqrt 2 \\
\Leftrightarrow \frac{1}{{\sqrt 2 }}\sqrt {{{\left( {\sqrt {2x - 1} + 1} \right)}^2}} + \frac{1}{{\sqrt 2 }}\sqrt {{{\left( {\sqrt {2x - 1} - 1} \right)}^2}} = \sqrt 2 \\
\Leftrightarrow \left| {\sqrt {2x - 1} + 1} \right| + \left| {\sqrt {2x - 1} - 1} \right| = 2\\
\Leftrightarrow \sqrt {2x - 1} + 1 + \left| {\sqrt {2x - 1} - 1} \right| = 2\\
TH1:\,\,\,\,\frac{1}{2} \le x \le 1 \Leftrightarrow 0 \le 2x - 1 \le 1 \Rightarrow \sqrt {2x - 1} - 1 \le 0\\
\Rightarrow \sqrt {2x - 1} + 1 + 1 - \sqrt {2x - 1} = 2\\
\Leftrightarrow 2 = 2\,\,\,\,\,\left( {\forall x} \right)\\
TH2:\,\,\,\,x > 1 \Rightarrow 2x - 1 > 1 \Rightarrow \sqrt {2x - 1} - 1 > 0\\
\Rightarrow \sqrt {2x - 1} + 1 + \sqrt {2x - 1} - 1 = 2\\
\Leftrightarrow \sqrt {2x - 1} = 1\\
\Leftrightarrow x = 1\,\,\,\left( {L,x > 1} \right)\\
\Rightarrow \frac{1}{2} \le x \le 1
\end{array}\)
