Tháng Hai 24, 2024

Tính: \(A=\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\) A 1 B 3 C 2 D 4

Tính: \(A=\frac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

A 1

B 3

C 2

D 4

Hướng dẫn Chọn đáp án là: A

Lời giải chi tiết:

Ta có

\(\begin{array}{l}A = \frac{{\sqrt 2 }}{{2\sqrt 2 + \sqrt {3 + \sqrt 5 } }} + \frac{{\sqrt 2 }}{{2\sqrt 2 – \sqrt {3 – \sqrt 5 } }}\\= \frac{{\sqrt 2 .\sqrt 2 }}{{2\sqrt 2 .\sqrt 2 + \sqrt 2 .\sqrt {3 + \sqrt 5 } }} + \frac{{\sqrt 2 .\sqrt 2 }}{{2\sqrt 2 .\sqrt 2 – \sqrt 2 .\sqrt {3 – \sqrt 5 } }}\\= \frac{2}{{4 + \sqrt {6 + 2\sqrt 5 } }} + \frac{2}{{4 – \sqrt {6 – 2\sqrt 5 } }}\\= \frac{2}{{4 + \sqrt {{{\sqrt 5 }^2} + 2\sqrt 5 + 1} }} + \frac{2}{{4 – \sqrt {{{\sqrt 5 }^2} – 2\sqrt 5 + 1} }}\\= \frac{2}{{4 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} }} + \frac{2}{{4 – \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} }}\\= \frac{2}{{4 + \sqrt 5 + 1}} + \frac{2}{{4 – \sqrt 5 + 1}}\\= \frac{2}{{5 + \sqrt 5 }} + \frac{2}{{5 – \sqrt 5 }}\\= \frac{{2.\left( {5 – \sqrt 5 } \right) + 2.\left( {5 + \sqrt 5 } \right)}}{{\left( {5 + \sqrt 5 } \right).\left( {5 – \sqrt 5 } \right)}}\\= \frac{{10 – 2\sqrt 5 + 10 + 2\sqrt 5 }}{{20}} = 1\end{array}\)