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

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

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

b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

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

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

\(=\sqrt{2}\)

c: \(\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)=5-4=1\)

c: Ta có: \(C=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\)

\(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\sqrt{10}\)

a: =2-căn 3-2-căn 3

=-2căn 3

b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{7}+1-\sqrt{7}+1\right)=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)

=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)

\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)

=>\(A=\sqrt{10+2\sqrt{5}}\)

b) làm thế nào để ra được \(\dfrac{1}{ \sqrt{2}}\)\((\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}})\) vậy ạ?????

a.\(\sqrt{\left(\sqrt{7}-1\right)^2}=\left|\sqrt{7}-1\right|=\sqrt{7}-1\)

b.\(\sqrt{\left(2-\sqrt{3}\right)^2}=\left|2-\sqrt{3}\right|=2-\sqrt{3}\)

c.\(\sqrt{\left(\sqrt{2}+5\right)^2}-\sqrt{2}=\left|\sqrt{2}+5\right|-\sqrt{2}=\sqrt{2}+5-\sqrt{2}=5\)

d.\(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{5}-6\right)^2}=\left|3+\sqrt{5}\right|+\left|\sqrt{5}-6\right|=3+\sqrt{5}+6-\sqrt{5}=9\)

a)\(=\sqrt{7}-1\)

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

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

d)\(=3+\sqrt{5}+\sqrt{5}+6=9\)

`a,\sqrt(3+2sqrt2)=\sqrt((sqrt2)^2+2.sqrt2 .1+1^2)=\sqrt((sqrt2+1)^2)=|sqrt2+1|=sqrt2+1`

`b,\sqrt(7+4sqrt3)=\sqrt((sqrt3)^2+2.\sqrt3 .2 +2^2)=\sqrt((sqrt3+2)^2)=|sqrt3+2|=sqrt3+2`

`c,sqrt(14-6sqrt5)=\sqrt((sqrt5)^2-2.\sqrt5 .3+3^2)=sqrt((sqrt5-3)^2)=|sqrt5-3|+3-sqrt5`

a: Ta có: \(A=\left(\dfrac{6+\sqrt{20}}{3+\sqrt{5}}+\dfrac{\sqrt{14}-\sqrt{2}}{\sqrt{7}-1}\right):\left(2+\sqrt{2}\right)\)

\(=\left(2+\sqrt{2}\right):\left(2+\sqrt{2}\right)\)

=1

b: Ta có: \(B=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-\dfrac{11}{2\sqrt{3}+1}\)

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

=1

a: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{4^2-\left(10+2\sqrt5\right)}\)

=>\(A^2=8+2\cdot\sqrt{6-2\sqrt5}=8+2\left(\sqrt5-1\right)=6+2\sqrt5\)

=>\(A^2=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)

b: \(2\sqrt{8\sqrt3}-2\cdot\sqrt{5\sqrt3}-3\cdot\sqrt{20\sqrt3}\)

\(=2\cdot2\sqrt{2\sqrt3}-2\cdot\sqrt{5\sqrt3}-3\cdot2\sqrt{5\sqrt3}\)

\(=4\sqrt{2\sqrt3}-2\cdot\sqrt{5\sqrt3}-6\sqrt{5\sqrt3}=4\sqrt{2\sqrt3}-8\sqrt{5\sqrt3}\)

a: \(\frac{\sqrt{3-\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}\)

\(=\frac{\sqrt{6-2\sqrt5}\cdot\left(3+\sqrt5\right)}{\sqrt{20}+\sqrt4}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}\)

\(=\frac{\left(\sqrt5-1\right)^{}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+\sqrt1\right)}=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}=1\)

b: \(\sqrt{8\sqrt3}-\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt3}-5\sqrt{2\sqrt3}+4\sqrt{\sqrt{64}\cdot\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\sqrt{8\sqrt3}\)

\(=-3\sqrt{2\sqrt3}+4\cdot2\sqrt{2\sqrt3}=5\sqrt{2\sqrt3}\)

c: \(\sqrt{2-\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)\)

\(=\frac{\sqrt{4-2\sqrt3}\cdot\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt3-1\right)\left(\sqrt5+\sqrt2\right)}{\sqrt2}=\frac{\left(\sqrt6-\sqrt2\right)\left(\sqrt5+\sqrt2\right)}{2}\)

\(=\frac{\sqrt{30}+2\sqrt3-\sqrt{10}-2}{2}\)

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

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

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

e: Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(A^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{16-\left(10+2\sqrt5\right)}\)

=>\(A^2=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(A^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=8+2\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(A=\sqrt5+1\)

f: \(\left(5+2\sqrt6\right)\left(49-20\sqrt6\right)\cdot\sqrt{5-2\sqrt6}\)

\(=\left(245-100\sqrt6+98\sqrt6-240\right)\cdot\sqrt{\left(\sqrt3-\sqrt2\right)^2}\)

\(=\left(5-2\sqrt6\right)\left(\sqrt3-\sqrt2\right)=\left(\sqrt3-\sqrt2\right)^2\)

g: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2-\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4-2\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2}{2-\sqrt{\left(\sqrt3-1\right)^2}}\)

\(=\frac{\sqrt2}{2+\left(\sqrt3+1\right)^{}}+\frac{\sqrt2}{2-\left(\sqrt3-1\right)}\)

\(=\frac{\sqrt2}{2+\sqrt3+1^{}}+\frac{\sqrt2}{2-\sqrt3+1}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{9-3}\)

\(=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{6}=\frac{6\sqrt2}{6}=\sqrt2\)

i: \(\frac{\left(\sqrt5+2\right)^2-8\sqrt5}{2\sqrt5-4}\)

\(=\frac{9+4\sqrt5-8\sqrt5}{2\left(\sqrt5-2\right)}\)

\(=\frac{9-4\sqrt5}{2\left(\sqrt5-2\right)}=\frac{\left(\sqrt5-2\right)^2}{2\left(\sqrt5-2\right)}=\frac{\sqrt5-2}{2}\)

k: \(\sqrt{14-8\sqrt3}-\sqrt{24-12\sqrt3}\)
\(=\sqrt{8-2\cdot2\sqrt2\cdot\sqrt6+6}-\sqrt{6\left(4-2\sqrt3\right)}\)
\(=\sqrt{\left(2\sqrt2-\sqrt6\right)^2}-\sqrt6\left(\sqrt3-1\right)=2\sqrt2-\sqrt6-\sqrt{18}+\sqrt6=2\sqrt2-3\sqrt2=-\sqrt2\)

l: \(\frac{4}{\sqrt3+1}+\frac{1}{\sqrt3-2}+\frac{6}{\sqrt3-3}\)

\(=\frac{4\left(\sqrt3-1\right)}{3-1}-\frac{1\left(2+\sqrt3\right)}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}-\frac{6\left(3+\sqrt3\right)}{9-3}\)

\(=2\left(\sqrt3-1\right)-\left(2+\sqrt3\right)-\left(3+\sqrt3\right)=2\sqrt3-2-2-\sqrt3-3-\sqrt3\)

=-7

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

\(=\left(2\sqrt2+3\cdot2\cdot1+3\cdot\sqrt2\cdot1+1\right)-\left(2\sqrt2-3\cdot2\cdot1+3\cdot\sqrt2\cdot1-1\right)\)

\(=\left(5\sqrt2+7\right)-\left(5\sqrt2-7\right)=14\)