4: Ta có: \(\dfrac{6}{1-\sqrt{3}}-\dfrac{3\sqrt{3}+3}{\sqrt{3}+1}\)

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

\(=-6-3\sqrt{3}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+2\sqrt{12}}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-2\sqrt{75}}}}\)

\(C=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(C=\sqrt{4+5}\)

\(C=3\)

a: \(=\dfrac{6+4\sqrt{2}}{\sqrt{2}+2+\sqrt{2}}+\dfrac{6-4\sqrt{2}}{\sqrt{2}-2+\sqrt{2}}\)

\(=\dfrac{6+4\sqrt{2}}{2+2\sqrt{2}}+\dfrac{6-4\sqrt{2}}{2\sqrt{2}-2}\)

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

=căn 2+1+căn 2-1=2căn 2

b: \(=\dfrac{\sqrt{3}+\sqrt{3+\sqrt{3}}+\sqrt{3}-\sqrt{3+\sqrt{3}}}{1-\sqrt{3}-1}=\dfrac{-2\sqrt{3}}{\sqrt{3}}=-2\)

bạn ơi cho mình hỏi câu b chi tiết hơn đước ko ạ

mình chưa hiểu lắm

Câu 1,2,3 Ez quá rồi :3

Câu 4:

Tổng quát:

\(\frac{1}{\sqrt{a}+\sqrt{a+1}}=\frac{\sqrt{a}-\sqrt{a+1}}{a-a-1}=\sqrt{a+1}-\sqrt{a}.\) Game là dễ :v

Câu 5 ko khác câu 4 lắm :v

Câu 5: 

Tổng quát:

\(\frac{1}{\sqrt{a}-\sqrt{a+1}}=\frac{\sqrt{a}+\sqrt{a+1}}{a-a-1}=-\sqrt{a}-\sqrt{a+1}.\) Game là dễ :v

1: \(\left(\sqrt{15}-2\sqrt3\right)^2+12\sqrt5\)

\(=15+12-2\cdot\sqrt{15}\cdot2\sqrt3+12\sqrt5\)

\(=27-4\sqrt{45}+12\sqrt5=27\)

2: \(3\sqrt2\left(4-\sqrt2\right)+3\left(1-2\sqrt2\right)^2\)

\(=12\sqrt2-6+3\left(9-4\sqrt2\right)\)

\(=12\sqrt2-6+27-12\sqrt2=21\)

3: \(\frac12\left(\sqrt6+\sqrt5\right)^2-\frac14\cdot\sqrt{120}-\sqrt{\frac{15}{2}}\)

\(=\frac12\left(11+2\sqrt{30}\right)-\frac14\cdot2\sqrt{30}-\sqrt{\frac{30}{4}}\)

\(=\frac{11}{2}+\sqrt{30}-\frac12\sqrt{30}-\frac12\sqrt{30}=\frac{11}{2}\)

4: \(\left(\sqrt{4-\sqrt7}-\sqrt{4+\sqrt7}\right)^2\)

\(=4-\sqrt7+4+\sqrt7-2\cdot\sqrt{\left(4-\sqrt7\right)\left(4+\sqrt7\right)}\)

\(=8-2\cdot\sqrt{16-7}=8-2\cdot\sqrt9=8-2\cdot3=8-6=2\)

5: \(\left(\sqrt{\sqrt{14}+\sqrt5}+\sqrt{\sqrt{14}-\sqrt5}\right)^2\)

\(=\sqrt{14}+\sqrt5+\sqrt{14}-\sqrt5+2\cdot\sqrt{\left(\sqrt{14}+\sqrt5\right)\left(\sqrt{14}-\sqrt5\right)}\)

\(=2\sqrt{14}+2\cdot\sqrt{14-5}=2\sqrt{14}+2\cdot\sqrt9=2\sqrt{14}+6\)

6: \(\left(\sqrt3+1\right)^3-\left(\sqrt3-1\right)^3\)

\(=\left(3\sqrt3+3\cdot3\cdot1+3\cdot\sqrt3\cdot1^2+1\right)-\left(3\sqrt3-3\cdot3\cdot1+3\sqrt3-1\right)\)

\(=\left(6\sqrt3+10\right)-\left(6\sqrt3-10\right)=20\)

7: \(\left(\sqrt2+1\right)^3-\left(\sqrt2-1\right)^3\)

\(=\left\lbrack\left(\sqrt2\right)^3+3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2+1^3\right\rbrack-\left\lbrack\left(\sqrt2\right)^3-3\cdot\left(\sqrt2\right)^2\cdot1+3\cdot\sqrt2\cdot1^2-1^3\right\rbrack\)

\(=\left(2\sqrt2+6+3\sqrt2+1\right)-\left(2\sqrt2-6+3\sqrt2-1\right)=5\sqrt2+7-5\sqrt2+7=14\)

8: \(\sqrt{13-\sqrt{160}}-\sqrt{53+4\sqrt{90}}\)

\(=\sqrt{13-4\sqrt{10}}-\sqrt{53+2\cdot\sqrt{360}}\)

\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt5+5}-\sqrt{45+2\cdot3\sqrt5\cdot2\sqrt2+8}\)

\(=\sqrt{\left(2\sqrt2-\sqrt5\right)^2}-\sqrt{\left(3\sqrt5+2\sqrt2\right)^2}\)

\(=2\sqrt2-\sqrt5-3\sqrt5-2\sqrt2=-4\sqrt5\)

9: \(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

\(=\frac{1}{\sqrt2}\left(\sqrt{6-2\sqrt5}+\sqrt{6+2\sqrt5}\right)\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt5-1\right)^2}+\sqrt{\left(\sqrt5+1\right)^2}\right)\)

\(=\frac12\left(\sqrt5-1+\sqrt5+1\right)=\frac{2\sqrt5}{\sqrt2}=\sqrt{10}\)

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{\dfrac{1}{2}\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{2}\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\dfrac{\sqrt{2}}{2}\left|\sqrt{3}-\sqrt{5}\right|+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left|\sqrt{5}-1\right|\)

\(=\dfrac{\sqrt{2}}{2}\left(\sqrt{5}-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}\left(\sqrt{3}+\sqrt{5}\right)-\sqrt{2}\left(\sqrt{5}-1\right)\)

\(=\sqrt{5}.\sqrt{2}-\sqrt{2}\left(\sqrt{5}-1\right)=\sqrt{2}\)

3.\(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{1-\left(\sqrt{5}\right)^2}\)

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)

\(=\dfrac{\left|\sqrt{3}-1\right|}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\left|\sqrt{3}-1\right|}=\dfrac{\sqrt{3}-1}{\sqrt{3}+1}+\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)

\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)\)

=-2

4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

=4

\(1,\left(2+\sqrt{3}\right)\left(7-4\sqrt{3}\right)\\ =14-8\sqrt{3}+7\sqrt{3}-12\\ =2-\sqrt{3}\\ 2,\left(\sqrt{5-2\sqrt{6}}+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\right)\sqrt{3}\\ =\left(\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\right)\sqrt{3}\\ =\left(\sqrt{3}-\sqrt{2}+\sqrt{2}\right)\sqrt{3}\\ =\sqrt{3}.\sqrt{3}\\ =3\\ 3,\sqrt{4+2\sqrt{3}}-\sqrt{5-2\sqrt{6}}+\sqrt{2}\\ =\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{2}\\ =\left|\sqrt{3}+1\right|-\left|\sqrt{3}-\sqrt{2}\right|+\sqrt{2}\\ =\sqrt{3}+1-\sqrt{3}-\sqrt{2}+\sqrt{2}\\ =1\\ 4,\sqrt{3+2\sqrt{2}}+\sqrt{6-4\sqrt{2}}\\ =\sqrt{\left(1+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{4}-\sqrt{2}\right)^2}\\ =\left|1+\sqrt{2}\right|+\left|\sqrt{4}-\sqrt{2}\right|\\ =1+\sqrt{2}+\sqrt{4}-\sqrt{2}\\ =1+\sqrt{4}\\ 5,2+\sqrt{17-4\sqrt{9+4\sqrt{5}}}\\ =2+\sqrt{17-8-4\sqrt{5}}\\ =2+\sqrt{\left(\sqrt{5}-2\right)^2}\\ =2+\left|\sqrt{5}-2\right|\\ =2+\sqrt{5}-2\\ =\sqrt{5}\)