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a) \(\left(\sqrt{\dfrac{9}{20}}-\sqrt{\dfrac{1}{2}}\right).\sqrt{2}=\sqrt{\dfrac{9}{20}.2}-\sqrt{\dfrac{1}{2}.2}=\sqrt{\dfrac{9}{10}}-1=\dfrac{3}{\sqrt{10}}-1\)
\(=\dfrac{3\sqrt{10}}{10}-1\)
b) \(\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)\sqrt{3}=\sqrt{12.3}+\sqrt{27.3}-\sqrt{3.3}\)
\(=\sqrt{36}+\sqrt{81}-\sqrt{9}=6+9-3=12\)
c) \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right)\sqrt{6}=\sqrt{\dfrac{8}{3}.6}-\sqrt{24.6}+\sqrt{\dfrac{50}{3}.6}\)
\(=\sqrt{16}-\sqrt{144}+\sqrt{100}=4-12+10=2\)
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) \(\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{2}\right|+\left|\sqrt{5}+\sqrt{2}\right|\)
\(=\sqrt{5}-\sqrt{2}+\sqrt{5}+\sqrt{2}\)
\(=\sqrt{5}+\sqrt{5}\)
\(=2\sqrt{5}\)
b) \(\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}\)
\(=\left|\sqrt{2}-1\right|-\left|\sqrt{2}-5\right|\)
\(=\sqrt{2}-1-\left(5-\sqrt{2}\right)\)
\(=\sqrt{2}-1-5+\sqrt{2}\)
\(=2\sqrt{2}-6\)
Trả lời:
\(A=\sqrt{3}-\frac{\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)
\(A=\sqrt{3}+\frac{\sqrt{6}}{\sqrt{2}-1}-\frac{2\sqrt{2}+2}{\sqrt{2}+1}\)
\(A=\sqrt{3}+\frac{\sqrt{6}.\left(\sqrt{2}+1\right)}{2-1}-\frac{2.\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
\(A=\sqrt{3}+\sqrt{6}.\left(\sqrt{2}+1\right)-2\)
\(A=\sqrt{3}+\sqrt{12}+\sqrt{6}-2\)
\(A=\sqrt{3}+2\sqrt{3}+\sqrt{6}-2\)
\(A=3\sqrt{3}+\sqrt{6}-2\)
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