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2/ Tìm y:
y + \(\dfrac{1}{3}\) x 4 = 12 \(\dfrac{2}{5}\) + y = \(\dfrac{12}{35}\) x \(\dfrac{14}{9}\)
\(y=12-\left(\dfrac{1}{3}\times4\right)=12-\dfrac{4}{3}=\dfrac{32}{3}\)
\(y=\left(\dfrac{12}{35}\times\dfrac{14}{9}\right)-\dfrac{2}{5}=\dfrac{8}{15}-\dfrac{2}{5}=\dfrac{2}{15}\)
2/3 = 2 x 4/ 3 x 4 =8/12
8/14 = 8 : 2/ 14 : 2 =4/7
5/9 = 5 x3/ 9 x 3 =15/37
35/ 40 = 35 : 5/ 40 : 5=7/8
\(\frac{18}{7}:9=\frac{18}{7}:\frac{9}{1}=\frac{18}{7}×\frac{1}{9}=\frac{9×2×1}{7×9}=\frac{2}{7}\)
\(\frac{64}{35}:4=\frac{64}{35}:\frac{4}{1}=\frac{64}{35}×\frac{1}{4}=\frac{16×4×1}{35×4}=\frac{16}{35}\)
\(\frac{24}{11}:12=\frac{24}{11}:\frac{12}{1}=\frac{24}{11}×\frac{1}{12}=\frac{12×2×1}{11×12}=\frac{2}{11}\)
\(\frac{35}{14}:5=\frac{35}{14}:\frac{5}{1}=\frac{35}{14}×\frac{1}{5}=\frac{5×7×1}{14×5}=\frac{7}{14}=\frac{1}{2}\)
18/17:9=18/17:9/1=18/17x1/9=2/17
64/35:4=64/35:4/1=64/35x1/4=16/35
24/11:12=24/11:12/1=24/11x1/12=2/11
35/4:5=35/4:5/1=35/4x1/5=7/4
A =\(\frac{2}{5}\times\frac{4}{7}+\frac{3}{5}\times\frac{4}{7}\) B = \(\frac{17}{13}\times\frac{12}{17}-\frac{4}{13}\times\frac{12}{17}\) C = \(\frac{9}{14}:\frac{6}{7}+\frac{5}{14}:\frac{6}{7}\) D =\(\frac{9}{12}:\frac{7}{9}\times\frac{12}{19}\)
= \(\frac{4}{7}\times\left(\frac{2}{5}+\frac{3}{5}\right)\) =\(\left(\frac{17}{13}-\frac{4}{13}\right)\times\frac{12}{17}\) = \((\frac{9}{14}+\frac{5}{14}):\frac{6}{7}\) = \(\frac{9}{12}\times\frac{9}{7}\times\frac{12}{19}\)
=\(\frac{4}{7}\times1\) = \(1\times\frac{12}{17}\) = \(1:\frac{6}{7}\) = \(\frac{9\times9\times12}{12\times7\times19}\)
=\(\frac{4}{7}\) = \(\frac{12}{17}\) =\(\frac{7}{6}\) = \(\frac{9\times9}{7\times19}=\frac{81}{133}\)
3/8
= 12/7 - 3/2
= 3/14