Question

[Phạm Mạnh Kiên]

có ai biết giải bài toán này k giúp mình với ?

1,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}+\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}}\)

2,\(\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)

3,\(\dfrac{3}{\sqrt{6}-\sqrt{3}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

4,\(\left(\sqrt{\dfrac{2}{3}-\sqrt{\dfrac{3}{2}}+\dfrac{5}{\sqrt{6}}}\right):\dfrac{6-\sqrt{6}}{1-\sqrt{6}}\)

5,\(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\times\left(\sqrt{3}+\sqrt{2}\right)\)

6,\(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\dfrac{\sqrt{5}+1}{\sqrt{5}-1}\)

7, \(\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)

8,\(\dfrac{4}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-2}+\dfrac{6}{\sqrt{3}-3}\)

9,\(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)

10,\(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+\dfrac{1}{\sqrt{4}+\sqrt{3}}\)

11,\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}+\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)

12,\(\dfrac{\sqrt{3}+2\sqrt{2}+\sqrt{3}-2\sqrt{2}}{\sqrt{3}+2\sqrt{2}-\sqrt{3}-2\sqrt{2}}\)

Answer 1

[Akai Haruma]

1. Sửa đề:

\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2+(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)

\(=\frac{2+\sqrt{3}+2-\sqrt{3}}{\sqrt{2^2-3}}=\frac{4}{1}=4\)

Answer 2

[Akai Haruma]

2.

\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)

\(=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=\frac{2\sqrt{3}}{1}=2\sqrt{3}\)