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a ) \(\left(-\frac{40}{51}.0,32.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(-\frac{40}{51}.\frac{8}{25}.\frac{17}{20}\right):\frac{64}{75}\)
\(=\left(\frac{-40.8.17}{51.25.20}\right):\frac{64}{75}\)
\(=\left(\frac{-16}{75}\right).\frac{75}{64}\)
\(=\frac{-1}{1}.\frac{1}{4}=-\frac{1}{4}\)
Ta có :
\(P=\frac{\frac{6}{8}+\frac{6}{10}+\frac{6}{14}+\frac{6}{26}}{\frac{11}{4}+\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(\Rightarrow P=\frac{\frac{3}{4}+\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
\(\Rightarrow P=\frac{3}{11}\)
Vậy \(P=\frac{3}{11}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}=\frac{3}{11}\)
đề bài của bn sai nên mk sửa luôn nha
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)
\(=\frac{11}{42}:\frac{1}{28}-8\)
\(=\frac{22}{3}-8\)
\(=-\frac{2}{3}\)
e, \(\frac{3}{14}:\frac{1}{28}-\frac{13}{21}:\frac{1}{28}+\frac{28}{42}:\frac{1}{28}-8\)
\(=\left(\frac{3}{14}-\frac{13}{21}+\frac{2}{3}\right):\frac{1}{28}-8\)
\(=\frac{11}{42}:\frac{1}{28}-8\)
\(=\frac{22}{3}-8\)
\(=-\frac{2}{3}\)