Đáp án:
d. \(D = \sqrt {11} \)
Giải thích các bước giải:
\(\begin{array}{l}
b.B = \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\
\to {B^2} = 4 + \sqrt {10 + 2\sqrt 5 } + 2\sqrt {{4^2} - \left( {10 + 2\sqrt 5 } \right)} + 4 - \sqrt {10 + 2\sqrt 5 } \\
= 8 + 2\sqrt {16 - 10 - 2\sqrt 5 } \\
= 8 + 2\sqrt {6 - 2\sqrt 5 } \\
= 8 + 2\sqrt {5 - 2.\sqrt 5 .1 + 1} \\
= 8 + 2\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= 8 + 2\left( {\sqrt 5 - 1} \right)\\
= 6 + 2\sqrt 5 \\
b.DK:x \ge 2\\
B = \sqrt {x - 2 + 2.\sqrt {x - 2} .\sqrt 2 + 2} + \sqrt {x - 2 - 2.\sqrt {x - 2} .\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt {x - 2} + \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt {x - 2} - \sqrt 2 } \right)}^2}} \\
= \sqrt {x - 2} + \sqrt 2 + \left| {\sqrt {x - 2} - \sqrt 2 } \right|\\
\to \left[ \begin{array}{l}
B = \sqrt {x - 2} + \sqrt 2 + \sqrt {x - 2} - \sqrt 2 \left( {DK:\sqrt {x - 2} \ge \sqrt 2 \to x \ge 4} \right)\\
B = \sqrt {x - 2} + \sqrt 2 - \sqrt {x - 2} + \sqrt 2 \left( {DK:2 \le x < 4} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
B = 2\sqrt {x - 2} \\
B = 2\sqrt 2
\end{array} \right.\\
c.DK:x \ge 1\\
C = \sqrt {x - 1 + 2\sqrt {x\left( {x - 1} \right)} + x} - \sqrt {x - 1 - 2\sqrt {x\left( {x - 1} \right)} + x} \\
= \sqrt {{{\left( {\sqrt {x - 1} + \sqrt x } \right)}^2}} - \sqrt {{{\left( {\sqrt {x - 1} - \sqrt x } \right)}^2}} \\
= \sqrt {x - 1} + \sqrt x - \left| {\sqrt {x - 1} - \sqrt x } \right|\\
= \sqrt {x - 1} + \sqrt x + \sqrt {x - 1} - \sqrt x \\
= 2\sqrt {x - 1} \\
d.Do:\sqrt {x - 3} + \sqrt {x + 3} = \sqrt {11} \\
\to x - 3 + 2\sqrt {{x^2} - 9} + x + 3 = 11\\
\to 2x + 2\sqrt {{x^2} - 9} = 11\\
\to \sqrt {2x + 2\sqrt {{x^2} - 9} } = \sqrt {11} \\
\to D = \sqrt {11}
\end{array}\)
