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a)\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
\(=2.\sqrt{\dfrac{4^2}{3}}-3.\sqrt{\dfrac{1}{3.3^2}}-6\sqrt{\dfrac{2^2}{3.5^2}}\)
\(=2.\dfrac{4}{\sqrt{3}}-3.\dfrac{1}{3\sqrt{3}}-6.\dfrac{2}{5\sqrt{3}}=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)\(=\dfrac{23}{5\sqrt{3}}=\dfrac{23\sqrt{3}}{15}\)
b)\(\left(6\sqrt{\dfrac{8}{9}}-5\sqrt{\dfrac{32}{25}}+14\sqrt{\dfrac{18}{49}}\right).\sqrt{\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{8}{9}.\dfrac{1}{2}}-5\sqrt{\dfrac{32}{25}.\dfrac{1}{2}}+14\sqrt{\dfrac{18}{49}.\dfrac{1}{2}}\)
\(=6\sqrt{\dfrac{4}{9}}-5\sqrt{\dfrac{16}{25}}+14\sqrt{\dfrac{9}{49}}\)\(=6.\dfrac{2}{3}-5.\dfrac{4}{5}+14.\dfrac{3}{7}=6\)
c)\(\sqrt{\left(\sqrt{2}-2\right)^2}-\sqrt{6+4\sqrt{2}}=\left|\sqrt{2}-2\right|-\sqrt{4+2.2\sqrt{2}+2}=2-\sqrt{2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)
\(=2-\sqrt{2}-\left(2+\sqrt{2}\right)=-2\sqrt{2}\)
a,
\(\sqrt{0,0004}=0.02\)
\(\sqrt{\frac{16}{81}}=\frac{\sqrt{16}}{\sqrt{81}}=\frac{4}{9}\)
\(\sqrt{25}=5\)
\(\sqrt{0,16}=0,4\)
b,\(\sqrt{\frac{9}{16}}+\sqrt{\frac{25}{9}}\)
= \(\frac{\sqrt{9}}{\sqrt{16}}+\frac{\sqrt{25}}{\sqrt{9}}\)
= \(\frac{3}{4}+\frac{5}{3}\)
=\(\frac{29}{12}\)
a)
Ta có :
\(\sqrt{0,16}\) + \(\sqrt{\frac{4}{25}}\) = \(\sqrt{\left(0,4^2\right)}\) + \(\sqrt{\left(\frac{2}{5}\right)^2}\) = 0,4 + \(\frac{2}{5}\) = \(\frac{2}{5}+\frac{2}{5}\) = \(\frac{4}{5}\)
b)
Ta có :
\(\sqrt{3\frac{3}{16}}\) - \(\sqrt{0,36}\) = \(\sqrt{\left(\frac{7}{4}\right)^2}\) - \(\sqrt{\left(0,6^2\right)}\) = \(\frac{7}{4}-\frac{3}{5}=\frac{23}{20}\)