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Câu 1:\(\frac{12}{23}\times\frac{9}{28}+\frac{11}{23}\times\frac{9}{28}\)
\(=\frac{9}{28}\times\left(\frac{12}{23}+\frac{11}{23}\right)\)
\(=\frac{9}{28}\times\frac{23}{23}\)
\(=\frac{9}{28}\times1\)
\(=\frac{9}{28}\)
Câu 2:
\(3+\frac{1}{9}+\frac{3}{9}+2+\frac{5}{9}+3+2\)
\(=\left(3+2+2+3\right)+\left(\frac{1}{9}+\frac{3}{9}+\frac{5}{9}\right)\)
\(=10+\frac{9}{9}\)
\(=10+1\)
\(=11\)
\(\frac{12}{23}x\frac{9}{28}+\frac{11}{23}x\frac{9}{28}=\)\(\left(\frac{12}{23}+\frac{11}{23}\right)x\frac{9}{28}\)\(=\) \(1x\frac{9}{28}=\frac{9}{28}\)
\(3+\frac{1}{9}+\frac{3}{9}+2+\frac{5}{9}+3+2\)\(=\) \(\left(\frac{1}{9}+\frac{3}{9}+\frac{5}{9}\right)+\left(3+2+3+2\right)\)\(=\) \(1+3+2+3+2=\)\(11\)
$\frac{2}{3} = \frac{{2 \times 3}}{{3 \times 3}} = \frac{6}{9}$
$\frac{3}{4} = \frac{{3 \times 3}}{{4 \times 3}} = \frac{9}{{12}}$
Vậy $\frac{2}{3} = \frac{6}{9}$ ; $\frac{3}{4} = \frac{9}{{12}}$
\(\dfrac{2}{3}=\dfrac{6}{9}\)
\(\dfrac{3}{4}=\dfrac{9}{12}\)
a)\(\frac{2}{3}:\frac{4}{9}+\frac{1}{3}:\frac{4}{9}\)
=\(\frac{4}{9}:\left(\frac{2}{3}+\frac{1}{3}\right)\)
=\(\frac{4}{9}:1=\frac{4}{9}\)
b
a) $\frac{1}{6} + \frac{2}{3} = \frac{1}{6} + \frac{4}{6} = \frac{{1 + 4}}{6} = \frac{5}{6}$
b) $\frac{3}{5} + \frac{9}{{20}} = \frac{{12}}{{20}} + \frac{9}{{20}} = \frac{{12 + 9}}{{20}} = \frac{{21}}{{20}}$
c) $\frac{3}{8} + \frac{3}{{24}} = \frac{9}{{24}} + \frac{3}{{24}} = \frac{{9 + 3}}{{24}} = \frac{{12}}{{24}} = \frac{1}{2}$
a) $\frac{2}{7}:\frac{4}{9} = \frac{2}{7} \times \frac{9}{4}$
b) $\frac{{12}}{{25}}:\frac{6}{5} = \frac{{12}}{{25}} \times \frac{5}{6}$
c) $\frac{5}{8}:3 = \frac{5}{8}:\frac{3}{1} = \frac{5}{8} \times \frac{1}{3}$
\(a,\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}\)
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) $\frac{7}{9} \times \frac{{15}}{{28}} \times \frac{9}{7} = (\frac{7}{9} \times \frac{9}{7}) \times \frac{{15}}{{28}} = 1 \times \frac{{15}}{{28}} = \frac{{15}}{{28}}$
b) $\frac{9}{{32}} \times \left( {\frac{2}{3} \times \frac{{14}}{{21}}} \right) = \frac{9}{{32}} \times \left( {\frac{2}{3} \times \frac{2}{3}} \right) = \frac{9}{{32}} \times \frac{4}{9} = \frac{{36}}{{288}} = \frac{1}{8}$
3/9 + 2/3 = 3/9 + 6/9 = 9/9 = 1
k mk nha mk k lai
\(\frac{3}{9}+\frac{2}{3}=\frac{3}{9}+\frac{6}{9}=\frac{9}{9}=1\)