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

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

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

=0

h) Ta có: \(\sqrt{\dfrac{3+\sqrt{5}}{\sqrt{3-\sqrt{5}}}}+\sqrt{\dfrac{3-\sqrt{5}}{\sqrt{3+\sqrt{5}}}}\)

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

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

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

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

\(=\dfrac{8+2\sqrt{15}+8-2\sqrt{15}}{2}\)

=8

nhân liên hợp lên là ra nha bạn! ('ω')

Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)

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

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

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

=5

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

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

\(=\dfrac{\left(3+\sqrt{5}\right)\left(\sqrt{5}+1\right)-\left(\sqrt{5}-1\right)\left(3-\sqrt{5}\right)}{4}\)

\(=\dfrac{3\sqrt{5}+3+5+\sqrt{5}-3\sqrt{5}+5+3-\sqrt{5}}{4}\)

\(=4\)

\(\frac{1}{\sqrt{2}-\sqrt{3}}-\frac{1}{\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{7}}\)

= \(-\sqrt{3}-\sqrt{2}+\frac{\sqrt{5}+\sqrt{3}}{2}-\frac{\sqrt{7}+\sqrt{5}}{2}\)

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

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

Chúc bạn học tốt !!!

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

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

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

\(\Rightarrow A=-\sqrt{2}\)

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

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

\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)

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

\(\Rightarrow B=\sqrt{2}\)

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

\(=\dfrac{\sqrt{2}\left(2+\sqrt{5}\right)\left(3-\sqrt{5}\right)+\sqrt{2}\left(2-\sqrt{5}\right)\left(3+\sqrt{5}\right)}{4}\)

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

\(=\sqrt{2}\cdot\dfrac{2}{4}=\dfrac{1}{\sqrt{2}}\)

1:

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

2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)

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

\(=\dfrac{20-6}{2}=7\)