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\(a,\frac{2}{5}+\frac{1}{6}+\frac{3}{5}\)
\(=\left(\frac{2}{5}+\frac{3}{5}\right)+\frac{1}{6}\)
\(=1+\frac{1}{6}\)
\(=\frac{7}{6}\)
Hok tốt!~
\(=\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+...+\frac{1}{10}\)có 9 p/số
\(=\frac{1}{10}.9=\frac{9}{10}\)
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a \(\frac{4}{18}+\frac{5}{13}+\frac{7}{9}+\frac{21}{13}\)
= \(\left(\frac{4}{18}+\frac{7}{9}\right)+\left(\frac{5}{13}+\frac{21}{13}\right)\)
= \(1+2\)
= \(3\)
b \(\frac{4}{3}-\frac{2}{7}-\frac{5}{7}+\frac{2}{3}\)
= \(\left(\frac{4}{3}-\frac{2}{3}\right)-\left(\frac{2}{7}+\frac{5}{7}\right)\)
= \(\frac{2}{3}-1\)
= \(-\frac{1}{3}\)
a) \(\frac{3}{5}+\frac{4}{7}+\frac{2}{5}+\frac{1}{7}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{4}{7}+\frac{1}{7}+\frac{2}{7}\right)\)
\(=\frac{5}{5}+\frac{7}{7}=1+1=2\)
a) \(\frac{3}{5}+\frac{4}{7}+\frac{2}{5}+\frac{1}{7}+\frac{2}{7}\)
\(=\left(\frac{3}{5}+\frac{2}{5}\right)+\left(\frac{4}{7}+\frac{1}{7}+\frac{2}{7}\right)\)
= 1 + 1
= 2
b) \(\frac{4}{9}+\frac{8}{9}+\frac{12}{9}+\frac{16}{9}+...+\frac{48}{9}+\frac{52}{9}+\frac{56}{9}\)
\(=\frac{4+8+12+16+...+48+52+56}{9}\)
Xét 4 + 8 + 12 + 16 + ... + 48 + 52 + 56
Số các số hạng là:
(56 - 4) : 4 + 1 = 14 (số)
4 + 8 + 12 + 16 + ... + 48 + 52 + 56 = (56 + 4) x 14 : 2 = 420
\(\frac{4+8+12+16+...+48+52+56}{9}=\frac{420}{9}=\frac{140}{3}\)
=\(\frac{3}{4}\times\left(\frac{8}{9}+\frac{2}{3}\right)\)
\(=\frac{3}{4}\times\frac{14}{9}\)
\(=\frac{7}{6}\)
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)\(\frac{8}{5}.\frac{4}{7}+\frac{4}{7}.\frac{2}{5}-1\)
\(=\frac{4}{7}.\left(\frac{8}{5}+\frac{2}{5}\right)-1\)
\(=\frac{4}{7}.2-1\)
\(=\frac{8}{7}-\frac{7}{7}=\frac{1}{7}\)
