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\(B=\frac{0,6-\frac{3}{11}+\frac{3}{13}}{1,4-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(B=\frac{\frac{3}{5}-\frac{3}{11}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(B=\frac{3\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(B=\frac{3}{5}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(B=\frac{3}{5}-\frac{2}{7}=\frac{11}{35}\)
\(C=\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{-\dfrac{5}{8}+\dfrac{5}{10}-\dfrac{5}{11}-\dfrac{5}{12}}+\dfrac{\dfrac{3}{2}+\dfrac{3}{3}-\dfrac{3}{4}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{4}}=\dfrac{-3}{5}+\dfrac{3}{5}=0\)
Huyen lop truog hoc gioi toan nhat lop khog biet la sao thanh vien trog lop biet lam
a)\(\frac{36^7}{2^{15}\cdot27^5}=\frac{36^7}{\left(2^3\right)^5\cdot27^5}\)
\(=\frac{36^7}{\left(8\cdot27\right)^5}=\frac{36^7}{216^5}\)
\(=\frac{36^7}{36^5\cdot6^5}=\frac{36^5\cdot36^2}{36^5\cdot6^5}\)
\(=\frac{36^2}{6^5}=\frac{\left(6^2\right)^2}{6^5}=\frac{6^4}{6^5}=\frac{1}{6}\)
\(\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}-\frac{-2}{9}-\frac{7}{5}\right)-\frac{5x}{2}\left(\frac{x}{5}-\frac{4}{5}\right)\)
\(M=\frac{-2x}{3}+3x\left(\frac{x}{6}+\frac{2}{9}-\frac{7}{5}\right)-\frac{5x}{2}.\frac{x-4}{5}\)
\(M=\frac{-2x}{3}+3x\left(\frac{15x+20-126}{90}\right)-\frac{5x^2-20x}{10}\)
\(M=\frac{-2x}{3}+3x.\frac{15x-106}{90}-\frac{5.\left(x^2-4x\right)}{10}\)
\(M=\frac{-2x}{3}+\frac{45x^2-318x}{90}-\frac{x^2-4x}{2}\)
Bài làm
\(C=-\frac{3}{5}.\frac{20}{135}.\frac{-7}{1400}.\frac{250}{3}.\frac{27}{10}\)
\(C=\frac{-3}{5}.\frac{4}{27}.\frac{-1}{200}.\frac{250}{3}.\frac{27}{10}\)
\(C=\frac{-3.4.\left(-1\right).250.27}{5.27.200.3.10}\)
\(C=\frac{1}{10}\)
Vậy \(C=\frac{1}{10}\)
# Học tốt #
C = \(\frac{-3}{5}\cdot\frac{20}{135}\cdot\frac{-7}{1400}\cdot\frac{250}{3}\cdot\frac{27}{10}\) C = \(\left(\frac{-3}{5}\cdot\frac{250}{3}\right)\cdot\left(\frac{20}{135}\cdot\frac{27}{10}\right)\cdot\frac{-7}{1400}\) C = \(-50\cdot\frac{2}{5}\cdot\frac{-7}{1400}\) C = \(-20\cdot\frac{-7}{1400}\) C = \(\frac{1}{10}\)