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a) \(\frac{3}{5}:\left(\frac{-1}{15}-\frac{1}{6}\right)+\frac{3}{5}:\left(\frac{-1}{3}-1\frac{1}{15}\right)\)
\(=\frac{3}{5}:\frac{-7}{30}+\frac{3}{5}:\frac{-7}{5}\)
\(=\frac{3}{5}\cdot\frac{30}{-7}+\frac{3}{5}\cdot\frac{5}{-7}\)
\(=\frac{3}{5}\left(\frac{-30}{7}+\frac{-5}{7}\right)=\frac{3}{5}\cdot-5=-3\)
b) \(10\cdot\sqrt{0,01}\cdot\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac{1}{6}\sqrt{4}\)
\(=10\cdot\frac{1}{10}\cdot\frac{4}{3}+3\cdot7-\frac{1}{6}\cdot2\)
\(=\frac{4}{3}+21-\frac{2}{6}=22\)
1. a) 3+2=5
b) 0,5-0,1=0,4
c) 4/5-1/9=31/45
d) 2-0,6=1,4
2. a) 8-4+3=7
b) 11+5-3=13
c) 3/2-4/6-7-37/6
d) 4+5-6=3
\(10.\sqrt{0,01}.\sqrt{\frac{16}{9}}+3\sqrt{19}-\frac{1}{6}\sqrt{4}\)
\(=10.\sqrt{\left(\frac{1}{10}\right)^2}.\sqrt{\left(\frac{4}{3}\right)^2}+3\sqrt{19}-\frac{1}{6}\sqrt{2^2}\)
\(=10.\frac{1}{10}.\frac{4}{3}+3\sqrt{19}-\frac{1}{6}.2\)
\(=\frac{4}{3}+3\sqrt{19}-\frac{1}{3}\)
\(=\left(\frac{4}{3}-\frac{1}{3}\right)+3\sqrt{19}\)
\(=1+3\sqrt{19}\)
1: \(=\left(\dfrac{1}{3}\right)^{25}\cdot90^{25}\cdot\dfrac{1}{3^{25}}-\dfrac{2}{12}\)
\(=\dfrac{30^{25}}{3^{25}}-\dfrac{1}{6}=10^{25}-\dfrac{1}{6}\)
2: \(=10\cdot1\cdot\dfrac{4}{3}+7-\dfrac{1}{6}\cdot2=20\)
\(=10.\dfrac{1}{10}.\dfrac{4}{3}+3.7-\dfrac{1}{6}.2\)
\(=1.\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{28}{21}+\dfrac{441}{21}+\dfrac{7}{21}\)
\(=\dfrac{469}{21}+\dfrac{7}{21}\)
\(=\dfrac{68}{3}\)
\(10.\sqrt{0,01}.\sqrt{\dfrac{16}{9}}+3.\sqrt{49}-\dfrac{1}{6}.\sqrt{4}\)
= 10 . 0,1 . \(\dfrac{4}{3}\) + 3. 7 - \(\dfrac{1}{6}.2\)
= 1 . \(\dfrac{4}{3}\) + 21 - \(\dfrac{1}{3}\)
= \(\dfrac{4}{3}+21-\dfrac{1}{3}\)
= 22
