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b: \(=\dfrac{\sqrt{5}+1}{\sqrt{5}-1}+\dfrac{\sqrt{5}-1}{\sqrt{5}+1}\)
\(=\dfrac{6+2\sqrt{5}+6-2\sqrt{5}}{4}=\dfrac{12}{4}=3\)
c: \(=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{13+30\left(\sqrt{2}+1\right)}=\sqrt{43+30\sqrt{2}}\)
e: \(=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}-\sqrt{2}}\)
\(=\dfrac{\sqrt{2}\cdot\sqrt{3+\sqrt{3}-1}}{\sqrt{3}-1}=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{\sqrt{3}-1}\)
\(=\dfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(A=\sqrt{9}-\sqrt{1}=3-1=2\)
1: \(=\sqrt{6}+\sqrt{6}+1=2\sqrt{6}+1\)
2: \(=\dfrac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
3: \(=\sqrt{3}+1-\sqrt{3}=1\)
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
a) \(\dfrac{30\left(\sqrt{6}-1\right)}{6-1}+\dfrac{2\left(\sqrt{6}+2\right)}{6-4}-\dfrac{6\left(3+\sqrt{6}\right)}{9-6}\)
= \(6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}=5\sqrt{6}+2\)
b) \(\dfrac{\sqrt{6+2\sqrt{5}}}{2}-\dfrac{\sqrt{5}-1}{2}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=1\)
c: \(=-\sqrt{3}+\dfrac{4+2\sqrt{3}}{2}=\dfrac{-2\sqrt{3}+4+2\sqrt{3}}{2}=\dfrac{4}{2}=2\)
a: \(=6\left(\sqrt{6}-1\right)+\sqrt{6}+2-6-2\sqrt{6}\)
\(=6\sqrt{6}-6-\sqrt{6}-4=5\sqrt{6}-10\)
b:\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}-\dfrac{\sqrt{5}-1}{2}\)
\(=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{2}=\dfrac{2}{2}=1\)
Lời giải:
a) Ta có:
\(14-6\sqrt{5}=14-2\sqrt{45}=9+5-2\sqrt{9.5}=(\sqrt{9}-\sqrt{5})^2=(3-\sqrt{5})^2\)
\(\Rightarrow \sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)
\(6+2\sqrt{5}=5+1+2\sqrt{5.1}=(\sqrt{5}+1)^2\)
\(\Rightarrow \sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)
Do đó: \(\sqrt{14-6\sqrt{5}}+\sqrt{6+2\sqrt{5}}=3-\sqrt{5}+\sqrt{5}+1=4\)
b)
\(\frac{\sqrt{10}+10}{1+\sqrt{10}}-\frac{5\sqrt{2}-2\sqrt{5}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{10}(1+\sqrt{10})}{1+\sqrt{10}}-\frac{\sqrt{10}(\sqrt{5}-\sqrt{2})}{\sqrt{5}-\sqrt{2}}\)
\(=\sqrt{10}-\sqrt{10}=0\)
\(E=\sqrt{\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+2\sqrt{6}}}=\sqrt{\dfrac{\left(5+2\sqrt{6}\right)^2}{5^2-\left(2\sqrt{6}\right)^2}}+\sqrt{\dfrac{\left(5-2\sqrt{6}\right)^2}{5^2-\left(2\sqrt{6}\right)^2}}=\sqrt{\dfrac{\left(5+2\sqrt{6}\right)^2}{25-24}}+\sqrt{\dfrac{\left(5-2\sqrt{6}\right)^2}{25-24}}=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}=5+2\sqrt{6}+5-2\sqrt{6}=5+5=10\)
\(E=\sqrt{\dfrac{5+2\sqrt{6}}{5-2\sqrt{6}}}+\sqrt{\dfrac{5-2\sqrt{6}}{5+2\sqrt{6}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2}{\left(\sqrt{3}-\sqrt{2}\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)
\(=\sqrt{\left(\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)^2}+\sqrt{\left(\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\right)^2}\)
\(=\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}+\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\)\(=\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\)
\(=\dfrac{3+2\sqrt{6}+2+2-2\sqrt{6}+3}{3-2}=10\)