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a: \(\frac{3}{4+\sqrt{9+4\sqrt5}}\)
\(=\frac{3}{4+\sqrt{\left(\sqrt5+2\right)^2}}\)
\(=\frac{3}{4+\sqrt5+2}=\frac{3}{6+\sqrt5}=\frac{3\left(6-\sqrt5\right)}{36-5}=\frac{3\left(6-\sqrt5\right)}{31}\)
b: \(\frac{\sqrt3}{\sqrt2+\sqrt{5+2\sqrt6}}\)
\(=\frac{\sqrt3}{\sqrt2+\sqrt{\left(\sqrt3+\sqrt2\right)^2}}=\frac{\sqrt3}{\sqrt2+\sqrt3+\sqrt2}\)
\(=\frac{\sqrt3}{2\sqrt2+\sqrt3}=\frac{\sqrt3\left(2\sqrt2-\sqrt3\right)}{8-3}=\frac{2\sqrt6-3}{5}\)
c: \(\frac{3}{\sqrt5+\sqrt7-\sqrt2}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{\left(\sqrt5+\sqrt7\right)^2-2}\)
\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)}{10+2\sqrt{35}}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(\sqrt{35}+5\right)\left(\sqrt{35}-5\right)}\)
\(=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{2\left(35-25\right)}=\frac{3\left(\sqrt5+\sqrt7+\sqrt2\right)\left(\sqrt{35}-5\right)}{20}\)
a: \(3=\sqrt9=\sqrt5\)
=>3+1>\(\sqrt5+1\)
=>\(4>\sqrt5+1\)
b: \(\sqrt{5+\sqrt7}<\sqrt{5+\sqrt{1936}}=\sqrt{5+44}=\sqrt{49}=7\)
a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
a: Ta có: \(4\sqrt{3a}-3\sqrt{12a}+\dfrac{6\sqrt{a}}{3}-2\sqrt{20a}\)
\(=4\sqrt{3a}-6\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)
\(=-2\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)
a: \(\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac14\cdot\sqrt8\right)\cdot3\sqrt6\)
\(=\left(2\sqrt6-4\sqrt3+5\sqrt2-\frac12\sqrt2\right)\cdot3\sqrt6\)
\(=\left(2\sqrt6-4\sqrt3+\frac92\cdot\sqrt2\right)\cdot3\sqrt6\)
\(=2\sqrt6\cdot3\sqrt6-4\sqrt3\cdot3\sqrt6+\frac92\cdot\sqrt2\cdot3\sqrt6\)
\(=36-12\sqrt{18}+\frac{27}{2}\sqrt{12}=36-36\sqrt2+27\sqrt3\)
b: \(\left(\sqrt{\frac17}-\sqrt{\frac{16}{7}}+\sqrt7\right):\sqrt7=\left(\frac{\sqrt7}{7}-\frac{4\sqrt7}{7}+\sqrt7\right):\sqrt7\)
\(=\frac17-\frac47+1=\frac87-\frac47=\frac47\)
c: \(\left(\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\right)^2\)
\(=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)
\(=6+2\cdot\sqrt{9-5}=6+2\cdot2=10\)
a,Ta có : \(1-\sqrt{3}\); \(\sqrt{2}-\sqrt{6}=\sqrt{2}\left(1-\sqrt{3}\right)\Rightarrow1-\sqrt{3}< \sqrt{2}\left(1-\sqrt{3}\right)\)
Vậy \(1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
b, Đặt A = \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)(*)
\(\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7}+1-\sqrt{7}+1-2=0\Rightarrow A=0\)
Vậy (*) = 0
1:
Ta có: \(\sqrt{2}-\sqrt{6}\)
\(=\sqrt{2}\left(1-\sqrt{3}\right)< 0\)
\(\Leftrightarrow1-\sqrt{3}< \sqrt{2}-\sqrt{6}\)
2:
Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2}{\sqrt{2}}\)
\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1-2}{\sqrt{2}}\)
=0