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\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)
a) $\frac{1}{6}:\frac{3}{7} = \frac{1}{6} \times \frac{7}{3} = \frac{7}{{18}}$
b) $\frac{5}{{12}}:\frac{1}{4} = \frac{5}{{12}} \times \frac{4}{1} = \frac{{5 \times 4}}{{12 \times 1}} = \frac{{5 \times 4}}{{4 \times 3 \times 1}} = \frac{5}{3}$
c) $\frac{4}{{15}}:\frac{8}{3} = \frac{4}{{15}} \times \frac{3}{8} = \frac{{4 \times 3}}{{15 \times 8}} = \frac{{4 \times 3}}{{5 \times 3 \times 4 \times 2}} = \frac{1}{{10}}$
d) $\frac{{18}}{5}:\frac{9}{{10}} = \frac{{18}}{5} \times \frac{{10}}{9} = \frac{{18 \times 10}}{{5 \times 9}} = \frac{{9 \times 2 \times 5 \times 2}}{{5 \times 9}} = 4$
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(S=1-\frac{1}{9}=\frac{8}{9}\)
\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)
\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)
\(=\frac{1}{24}\)
\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)
\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)
\(=\frac{1}{10}\)
\(a,\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}\)
Cách 1: \(\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}=1\times\frac{3}{7}=\frac{10}{7}\)
Cách 2: \(\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}=\frac{6}{11}\times\frac{3}{7}+\frac{5}{11}\times\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)
\(b,\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}\)
Cách 1: \(\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}=\frac{7}{15}+\frac{2}{15}=\frac{9}{15}\)
Cách 2: \(\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}=\frac{3}{5}\times1=\frac{3}{5}\)
P/s: Ý B có vấn đề thì phải
\(a,\left(\frac{6}{11}+\frac{5}{11}\right)x\frac{3}{7}\)
\(=\frac{11}{11}=1x\frac{3}{7}\)
\(=\frac{3}{7}\)
cach 2
\(\frac{6}{11}x\frac{3}{7}+\frac{5}{11}x\frac{3}{7}\)
\(=\frac{18}{77}+\frac{15}{77}\)
\(=\frac{33}{77}=\frac{3}{7}\)
\(b,\frac{3}{5}x\frac{7}{9}+\frac{3}{5}x\frac{2}{9}\)
\(=\frac{21}{45}+\frac{6}{45}\)
\(=\frac{27}{45}=\frac{3}{5}\)
cách 2 :
\(\frac{3}{5}x\left(\frac{7}{9}+\frac{2}{9}\right)\)
\(=\frac{3}{5}x\frac{9}{9}\)
\(=\frac{27}{45}=\frac{3}{5}\)
a) $\frac{1}{6} \times \frac{2}{3} = \frac{{1 \times 2}}{{6 \times 3}} = \frac{2}{{18}} = \frac{1}{9}$
b) $\frac{6}{5} \times \frac{3}{8} = \frac{{6 \times 3}}{{5 \times 8}} = \frac{{18}}{{40}} = \frac{9}{{20}}$
c) $\frac{4}{3} \times \frac{8}{9} = \frac{{4 \times 8}}{{3 \times 9}} = \frac{{32}}{{27}}$
d) $\frac{5}{{12}} \times \frac{{12}}{5} = \frac{{5 \times 12}}{{12 \times 5}} = \frac{{60}}{{60}} = 1$
a Đ
b S
c S
d Đ
a ) S
b ) Đ
c ) S
d ) Đ
k cho mk nhé