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\(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)có 9 p/số
\(=\frac{1}{10}.9=\frac{9}{10}\)
\(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}\)
\(=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\)
\(=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{13}{5}+\frac{12}{5}+\frac{3}{5}\right)\)
\(=1+\frac{28}{5}\)
\(=\frac{33}{5}\)
Ta có:
a) \(\frac{10}{18}+\frac{4}{9}+\frac{26}{10}+\frac{12}{5}+\frac{9}{15}=\frac{5}{9}+\frac{4}{9}+\frac{13}{5}+\frac{12}{5}+\frac{9}{15}=1+1+\frac{9}{15}=1\frac{9}{15}\)
b)\(\frac{10}{18}+\frac{4}{9}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}=\left(\frac{5}{9}+\frac{4}{9}\right)+\left(\frac{16}{128}+\frac{8}{128}+\frac{4}{128}+\frac{2}{128}+\frac{1}{128}\right)\)
\(=1+\frac{31}{128}=1\frac{31}{128}\)
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\(\frac{1}{2}\)x \(\frac{9}{10}\)+ \(\frac{8}{10}\)+ \(\frac{2}{4}\)+ \(\frac{3}{10}\)x \(0,5\)
= \(0,5\)x \(0,9\)+ \(0,8\)+ \(0,5\)x \(0,5\)
\(=0,45\)x \(1,3\)x \(0,5\)
\(=0,45\)x \(0,65=0,2925\)
a)\(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{1}{3}\right)+3\)
\(=1+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3\)
\(=1+3+\frac{7}{9}\)
\(=4\frac{7}{9}\)
\(=\frac{43}{9}\)
b)\(\frac{8}{5}\cdot\frac{4}{7}+\frac{4}{7}\cdot\frac{2}{5}-1\)
\(=\frac{4}{7}\cdot\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}\cdot2-1\)
\(=\frac{8}{7}-1\)
\(=\frac{1}{7}\)
a) \(\frac{8}{13}+\frac{4}{9}+\frac{1}{3}+\frac{5}{13}+3\)
\(=\left(\frac{8}{13}+\frac{5}{13}\right)+\left(\frac{4}{9}+\frac{3}{9}\right)+3\)
\(=1+\frac{7}{9}+3=4\frac{7}{9}\)
b)\(\frac{8}{5}.\frac{4}{7}+\frac{4}{7}.\frac{2}{5}-1\)
\(=\frac{4}{7}.\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}.2-1\)
\(=\frac{8}{7}-\frac{7}{7}=\frac{1}{7}\)