10×√0,01×√16/9+3√49-1/6√4
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8 tháng 6 2022
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: \(=10\cdot1\cdot\dfrac{4}{3}+7-\dfrac{1}{6}\cdot2=20\)
2 tháng 12 2017
\(=10.\dfrac{1}{10}.\dfrac{4}{3}+3.7-\dfrac{1}{6}.2\)
\(=1.\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{4}{3}+21+\dfrac{1}{3}\)
\(=\dfrac{28}{21}+\dfrac{441}{21}+\dfrac{7}{21}\)
\(=\dfrac{469}{21}+\dfrac{7}{21}\)
\(=\dfrac{68}{3}\)
2 tháng 12 2017
\(10.\sqrt{0,01}.\sqrt{\dfrac{16}{9}}+3.\sqrt{49}-\dfrac{1}{6}.\sqrt{4}\)
= 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
15 tháng 3 2020
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) \(10\cdot\sqrt{0,01}\cdot\sqrt{\frac{16}{9}}+3\sqrt{49}-\frac{1}{6}\sqrt{4}\)
\(=10\cdot\frac{1}{10}\cdot\frac{4}{3}+3\cdot7-\frac{1}{6}\cdot2\)
\(=\frac{4}{3}+21-\frac{2}{6}=22\)
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Nguyễn Thị Thương Hoài
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27 tháng 7 2024
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
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27 tháng 7 2024
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
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Nguyễn Thị Thương Hoài
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21 tháng 9 2024
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
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Nguyễn Thị Thương Hoài
Giáo viên
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21 tháng 9 2024
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
\(10\cdot\sqrt{0,01}\cdot\sqrt{\dfrac{16}{9}}+3\sqrt{49}-\dfrac{1}{6}\sqrt{4}\)
\(=10\cdot0,1\cdot\dfrac{4}{3}+3\cdot7-\dfrac{1}{6}\cdot2\)
\(=\dfrac{4}{3}+21-\dfrac{1}{3}=1+21=22\)
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
= \(\dfrac{4}{3}\) + 21 - \(\dfrac{1}{3}\)
= \(\left(\dfrac{4}{3}-\dfrac{1}{3}\right)\) + 21
= 1 + 21
= 22