Đáp án:
\(\begin{array}{l}
\left[ \begin{array}{l}
C = 2\sqrt x \\
C = 2
\end{array} \right.\\
\left[ \begin{array}{l}
D = 2\sqrt {x - 1} \\
D = 2
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C = \sqrt {x - 2\sqrt x + 1} + \sqrt {x + 2\sqrt x + 1} \\
= \sqrt {{{\left( {\sqrt x - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt x + 1} \right)}^2}} \\
= \left| {\sqrt x - 1} \right| + \sqrt x + 1\\
\to \left[ \begin{array}{l}
C = \sqrt x - 1 + \sqrt x + 1\left( {DK:x \ge 1} \right)\\
C = - \sqrt x + 1 + \sqrt x + 1\left( {DK:1 > x \ge 0} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
C = 2\sqrt x \\
C = 2
\end{array} \right.\\
D = \sqrt {x - 1 + 2\sqrt {x - 1} .1 + 1} + \sqrt {x - 1 - 2\sqrt {x - 1} .1 + 1} \\
= \sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} \\
= \sqrt {x - 1} + 1 + \left| {\sqrt {x - 1} - 1} \right|\\
\to \left[ \begin{array}{l}
D = \sqrt {x - 1} + 1 + \sqrt {x - 1} - 1\left( {DK:x \ge 2} \right)\\
D = \sqrt {x - 1} + 1 - \sqrt {x - 1} + 1\left( {DK:2 > x \ge 1} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
D = 2\sqrt {x - 1} \\
D = 2
\end{array} \right.
\end{array}\)
