Tính nhanh
A=\(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
B=\(\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}+\frac{5+\frac{5}{13}-\frac{5}{169}+\frac{5}{91}}{7+\frac{8}{13}-\frac{7}{169}}\)
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ta có: \(M=\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\)
\(M=\frac{7.9+7.2.9.3+7.3.9.4}{21.27+21.227.3+21.3.27.4}\)
\(M=\frac{7.9.\left(1+2.3+3.4\right)}{21.27.\left(1+2.3+3.4\right)}\)
\(M=\frac{7.9}{3.7.3.9}\)
\(M=\frac{1}{3^2}\)
\(M=\frac{1}{9}\)
và \(N=\frac{37}{333}=\frac{37}{37.9}=\frac{1}{9}\)
\(N=\frac{1}{9}\)
\(\Rightarrow M=N\left(=\frac{1}{9}\right)\)
M= \(\frac{7.9+14.27+21.36}{21.27+42.81+63.108}\).
M= \(\frac{7.9+14.27+21.36}{7.3.9.3+14.3.27.3+21.3+36.3}\).
M= \(\frac{7.9+14.27+21.36}{7.9.9+14.27.9+21.36.9}\).
M= \(\frac{7.9+14.27+21.36}{9\left(7.9+14.27+21.36\right)}\).
M= \(\frac{1}{9}\).
Ta có: N= \(\frac{37}{333}\)= \(\frac{1}{9}\).
Vì \(\frac{1}{9}\)= \(\frac{1}{9}\) nên M= N.
Vậy M= N.
a: \(=\dfrac{3\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}{4\left(\dfrac{1}{41}-\dfrac{4}{47}+\dfrac{9}{53}\right)}+\dfrac{-\dfrac{1}{4}\cdot\dfrac{-2}{3}-\dfrac{3}{4}:\dfrac{1}{6}}{\dfrac{3}{2}\cdot\left(\dfrac{-2}{3}-\dfrac{3}{4}\cdot\dfrac{-2}{3}\right)}\)
\(=\dfrac{3}{4}+\dfrac{\dfrac{2}{12}-\dfrac{9}{2}}{\dfrac{3}{2}\cdot\dfrac{-1}{6}}=\dfrac{3}{4}+\dfrac{-13}{3}:\dfrac{-3}{12}\)
\(=\dfrac{3}{4}+\dfrac{13}{3}\cdot\dfrac{12}{3}=\dfrac{3}{4}+\dfrac{156}{9}=\dfrac{217}{12}\)
b: \(A=158\left(\dfrac{12\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}{4\left(1-\dfrac{1}{7}-\dfrac{1}{289}-\dfrac{1}{85}\right)}:\dfrac{5\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}{6\left(1+\dfrac{1}{13}+\dfrac{1}{169}+\dfrac{1}{91}\right)}\right)\cdot\dfrac{50550505}{711711711}\)
\(=158\cdot\left(3\cdot\dfrac{6}{5}\right)\cdot\dfrac{50550505}{711711711}\)
\(\simeq40.39\)
\(B=\frac{0,6-\frac{3}{11}+\frac{3}{13}}{1,4-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(B=\frac{\frac{3}{5}-\frac{3}{11}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(B=\frac{3\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(B=\frac{3}{5}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(B=\frac{3}{5}-\frac{2}{7}=\frac{11}{35}\)
\(B=81.\left[\frac{3\left(12-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}\right)}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{79.2.1001001}{79.9.1001001}\)
\(B=81.\left[3.\frac{6}{5}\right].\frac{2}{9}\)
\(B=\frac{9.9.3.6.2}{5.9}\)
\(B=\frac{9.3.6.2}{5}\)
\(B=\frac{324}{5}\)
Tick cho minh nha Quang Hai Duong tick minh may man ca nam
\(A=\frac{7\times9+14\times27+21\times36}{21\times27+42\times81+63\times108}=\frac{7\times9+7\times2\times9\times3+7\times3\times9\times4}{21\times27+21\times2\times27\times3+21\times3\times27\times4}=\frac{7\times9\times\left(1+2\times3+3\times4\right)}{21\times27\times\left(1+2\times3\times3\times4\right)}=\frac{7\times3\times3}{7\times3\times3\times9}=\frac{1}{9}\)