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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: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)
=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)
\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)
=>\(A=\sqrt{10+2\sqrt{5}}\)
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(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: Ta có: \(\sqrt{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}+\sqrt{6}}{2}\)
b: Ta có: \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{2}}\)
\(=\left(3-\sqrt{2}\right)\left(3+\sqrt{2}\right)\)
=9-2
=7
c: Ta có: \(\left(\sqrt{7}+\sqrt{5}\right)\cdot\sqrt{12-2\sqrt{35}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=2
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: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)