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Question

rút gọn:

$A=\dfrac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}$

Akai Haruma · Accepted Answer

Lời giải:

Ta có:

$A\sqrt{2}=\frac{6+2\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}$

$=\frac{6+2\sqrt{5}}{2+\sqrt{5+1+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{5+1-2\sqrt{5}}}$

$=\frac{6+2\sqrt{5}}{2+\sqrt{(\sqrt{5}+1)^2}}+\frac{6-2\sqrt{5}}{2-\sqrt{(\sqrt{5}-1)^2}}$

$=\frac{2(3+2\sqrt{5})}{3+2\sqrt{5}}+\frac{2(3-2\sqrt{5})}{3-2\sqrt{5}}=2+2=4$

$\Rightarrow A=\frac{4}{\sqrt{2}}=2\sqrt{2}$Câu trả lời: Lời giải:

Ta có:

$A\sqrt{2}=\frac{6+2\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}$

$=\frac{6+2\sqrt{5}}{2+\sqrt{5+1+2\sqrt{5}}}+\frac{6-2\sqrt{5}}{2-\sqrt{5+1-2\sqrt{5}}}$

$=\frac{6+2\sqrt{5}}{2+\sqrt{(\sqrt{5}+1)^2}}+\frac{6-2\sqrt{5}}{2-\sqrt{(\sqrt{5}-1)^2}}$

$=\frac{2(3+2\sqrt{5})}{3+2\sqrt{5}}+\frac{2(3-2\sqrt{5})}{3-2\sqrt{5}}=2+2=4$

$\Rightarrow A=\frac{4}{\sqrt{2}}=2\sqrt{2}$

Violympic toán 9