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a, \(\sqrt{27.98}=\sqrt{3^3.49.2}=21\sqrt{6}\)
b, \(\sqrt{125.96}=\sqrt{5^3.4^2.6}=20\sqrt{30}\)
c, \(\sqrt{500.162}=\sqrt{10^2.5.9^2.2}=90\sqrt{10}\)
d, \(\sqrt{a^4b^5}=a^2b^2\sqrt{b}\)
a) \(\sqrt{\left(1-\sqrt{5}\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}=\left(\sqrt{5}-1\right)-\left(3-\sqrt{5}\right)=2\sqrt{5}-4\)
b) \(\frac{a-2\sqrt{a}+1}{\sqrt{a}-1}=\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}-1}=\sqrt{a}-1\) ( \(a\ge0\ne1\))
c) \(\frac{a+\sqrt{a}}{a}=\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}.\sqrt{a}}=\frac{\sqrt{a}+1}{\sqrt{a}}=1+\frac{1}{\sqrt{a}}\)(\(a>0\))
d) \(\frac{3+\sqrt{3}}{1+\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=\sqrt{3}\)
a: \(=2\cdot3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
b: \(=5\sqrt{10}+2\cdot5-5\sqrt{10}=10\)
c: \(=2\sqrt{7}\cdot\sqrt{7}-\sqrt{12}\cdot\sqrt{7}-\sqrt{7}\cdot\sqrt{7}+2\sqrt{21}=2\cdot7-7=7\)
d: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}=2\cdot11=22\)
a) \(\sqrt{\left(5-\sqrt{3}\right)^2}=\left|5-\sqrt{3}\right|=5-\sqrt{3}\)
b) \(\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=-\left(1-\sqrt{2}\right)=\sqrt{2}-1\)( vì 1 < √2 )
c) \(\sqrt{\left(\sqrt{3}-2\right)^2}=\left|\sqrt{3}-2\right|=-\left(\sqrt{3}-2\right)=2-\sqrt{3}\)( vì √3 < 2 )
\(2.3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
\(5\sqrt{10}+2.5-5\sqrt{10}=10\)
\(14-2\sqrt{21}-7+2\sqrt{21}=7\)
\(33-3\sqrt{22}-11+3\sqrt{22}=22\)