Question

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

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

2.

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

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

 

 

Answer 3

3.

\(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}=\frac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}+\frac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\)

\(=\frac{3(\sqrt{6}+\sqrt{3})}{3}+\frac{4(\sqrt{7}-\sqrt{3})}{4}=\sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}=\sqrt{6}+\sqrt{7}\)

 

Answer 4

4.

\(=\sqrt{\frac{2}{3}-\sqrt{\frac{3}{2}}+\frac{5}{\sqrt{6}}}:\frac{-\sqrt{6}(1-\sqrt{6})}{1-\sqrt{6}}=\sqrt{\frac{2}{3}-\sqrt{\frac{3}{2}}+\frac{5}{\sqrt{6}}}:(-\sqrt{6})\)

\(=-\sqrt{\frac{2}{3}-\sqrt{\frac{3}{2}}+\frac{5}{\sqrt{6}}}.\frac{1}{\sqrt{6}}=-\sqrt{\frac{1}{9}-\frac{\sqrt{6}}{12}+\frac{5\sqrt{6}}{36}}\)

\(=-\sqrt{\frac{1}{9}+\frac{\sqrt{6}}{18}}=-\sqrt{\frac{2+\sqrt{6}}{18}}=-\frac{\sqrt{4+2\sqrt{6}}}{6}\)

Answer 5

5.

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

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

Answer 6

6.

\(=2.\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)

\(=2.\frac{(\sqrt{5}+\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}-\frac{(\sqrt{5}+1)^2}{(\sqrt{5}-1)(\sqrt{5}+1)}\)

\(=\frac{2(\sqrt{5}+\sqrt{3})^2}{5-3}-\frac{(\sqrt{5}+1)^2}{5-1}\)

\(=(\sqrt{5}+\sqrt{3})^2-\frac{(\sqrt{5}+1)^2}{4}=8+2\sqrt{15}-\frac{3+\sqrt{5}}{2}\)

Answer 7

7.

\(=\frac{(\sqrt{5}+\sqrt{3})^2+(\sqrt{5}-\sqrt{3})^2}{(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})}=\frac{8+2\sqrt{15}+8-2\sqrt{15}}{5-3}=\frac{16}{2}=8\)

Answer 8

9.

,\(=\frac{4+3\sqrt{2}-(4-3\sqrt{2})}{(4-3\sqrt{2})(4+3\sqrt{2})}=\frac{6\sqrt{2}}{16-(3\sqrt{2})^2}=-3\sqrt{2}\)

Answer 9

10.

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

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

 

Answer 10

11.

\(=\left[\frac{-\sqrt{7}(1-\sqrt{2})}{1-\sqrt{2}}+\frac{\sqrt{5}(\sqrt{3}+1)^2}{(1-\sqrt{3})(1+\sqrt{3})}\right]:\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=(-\sqrt{7}-\sqrt{15}-2\sqrt{5}).(\sqrt{7}-\sqrt{5})\)

\(=-[(\sqrt{7}+\sqrt{5})+(\sqrt{15}+\sqrt{5})](\sqrt{7}-\sqrt{5})\)

\(=-[(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})+(\sqrt{15}+\sqrt{5})(\sqrt{7}-\sqrt{5})]\)

\(=-(2+\sqrt{105}-5\sqrt{3}+\sqrt{35}-5)=-(\sqrt{105}-5\sqrt{3}+\sqrt{35}-3)\)

Answer 11

12.

Mẫu số bằng $0$ nên phân số không xác định.

Answer 12

8.

\(=\frac{4(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}+\frac{\sqrt{3}+2}{(\sqrt{3}-2)(\sqrt{3}+2)}+\frac{6(\sqrt{3}+3)}{(\sqrt{3}-3)(\sqrt{3}+3)}\)

\(=\frac{4(\sqrt{3}-1)}{3-1}+\frac{\sqrt{3}+2}{3-4}+\frac{6(\sqrt{3}+3)}{3-9}\)

\(=2(\sqrt{3}-1)-(\sqrt{3}+2)-(\sqrt{3}+3)=-7\)