Đáp án:
g. \(\sqrt 2 \)
Giải thích các bước giải:
\(\begin{array}{l}
g.G = \sqrt {3 + \sqrt 5 } + \sqrt {7 - 3\sqrt 5 } - \sqrt 2 \\
= \dfrac{{\sqrt {6 + 2\sqrt 5 } + \sqrt {14 - 6\sqrt 5 } - 2}}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {5 + 2\sqrt 5 .1 + 1} + \sqrt {9 - 2.3.\sqrt 5 + 5} - 2}}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} + \sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} - 2}}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 5 + 1 + 3 - \sqrt 5 - 2}}{{\sqrt 2 }}\\
= \dfrac{2}{{\sqrt 2 }} = \sqrt 2 \\
h.DK:x \ge 2\\
H = \sqrt {x + 2\sqrt {2x - 4} } + \sqrt {x - 2\sqrt {2x - 4} } \\
= \dfrac{{\sqrt {2x + 4\sqrt {2x - 4} } + \sqrt {2x - 4\sqrt {2x - 4} } }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {2x - 4 + 2.2.\sqrt {2x - 4} + 4} + \sqrt {2x - 4 - 2.2.\sqrt {2x - 4} + 4} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt {2x - 4} + 2} \right)}^2}} + \sqrt {{{\left( {\sqrt {2x - 4} - 2} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {2x - 4} + 2 + \sqrt {2x - 4} - 2}}{{\sqrt 2 }}\left( {DK:\sqrt {2x - 4} \ge 2 \to 2x - 4 \ge 4 \to x \ge 4} \right)\\
= \dfrac{{2\sqrt {2x - 4} }}{{\sqrt 2 }} = \sqrt {4x - 8}
\end{array}\)
