Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{1}{3}-0,25+0,2}{\frac{7}{6}-0,875+0,7}+\frac{6}{7}\)
\(=\frac{1}{2}.\frac{2}{7}+\frac{6}{7}=1\)
\(A=\frac{\frac{1}{3}-\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}-\frac{2}{7}-\frac{2}{13}}.\frac{\frac{1}{3}-0,25+0,2}{\frac{7}{6}-0,875+0,7}+\frac{6}{7}\)
\(=\frac{1}{2}-\frac{1}{2}-\frac{1}{2}.\frac{5}{6}-\frac{25}{\frac{100}{-\frac{875}{1000}}}+\frac{6}{7}\)
\(=-\frac{1}{2}.\frac{5}{6}-\frac{25}{\frac{100}{\frac{-87,5}{100}}}+\frac{6}{7}\)
\(=-\frac{1}{2}.\frac{5}{6}-\frac{25}{-87,5}+\frac{6}{7}\)
đến đây tự lm ==
Ta có: \(A=\frac{\frac{3}{11}+1-\frac{3}{7}}{3+\frac{9}{11}-\frac{9}{7}}-\frac{\frac{1}{3}+0,25-\frac{1}{5}+0,125}{\frac{7}{6}+\frac{7}{8}-0,7+\frac{7}{16}}\)
\(=\frac{3\left(\frac{1}{11}+\frac{1}{3}-\frac{1}{7}\right)}{9\left(\frac{1}{3}+\frac{1}{11}-\frac{1}{7}\right)}-\frac{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}+\frac{1}{16}\right)}\)
\(=\frac{3}{9}-\frac{2}{7}=\frac{1}{3}-\frac{2}{7}=\frac{7}{21}-\frac{6}{21}=\frac{1}{21}\)
Vậy \(A=\frac{1}{21}\)
\(2014:\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1\frac{2}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{1\frac{1}{6}+0,875-0,7}{\frac{1}{3}+0,25-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}\right)\)
\(=2014:\left(\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\cdot\frac{\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}+\frac{2}{8}-\frac{2}{10}}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}{2\left(\frac{1}{6}+\frac{1}{8}-\frac{1}{10}\right)}\right)\)
\(=2014:\left(\frac{2}{7}\cdot\frac{7}{2}\right)=2014\)
\(Q=2002:\left[\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{-\frac{7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right]=2002:\left[\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}.\frac{-\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right]=2002:\left[\frac{2}{7}.\frac{-7}{2}\right]=2002.\left(-1\right)=-2002\)
\(Q=\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{\frac{-7}{6}-\frac{-7}{8}+\frac{-7}{10}}{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}\)
=>\(Q=\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}.\frac{-7.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{2.\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
=>\(Q=\frac{2}{7}.\frac{-7}{2}=\frac{2.7.\left(-1\right)}{7.2}=-1\)
=>Q=-1
\(M=\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2014}{2015}\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\frac{2014}{2015}\)
\(=\left[\frac{2.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7.\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right].\frac{2015}{2014}\)
\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right).\frac{2015}{2014}=\left(\frac{2}{7}-\frac{2}{7}\right).\frac{2015}{2014}=0\)
Ta có M = \(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2014}{2015}\)
= \(\left(\frac{2\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}{7\left(0,2-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-0,25+\frac{1}{5}\right)}\right):\frac{2014}{2015}\)
= \(\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\frac{2014}{2015}=\left(\frac{2}{7}-\frac{2}{7}\right):\left(\frac{2014}{2015}\right)=0\)
\(B=\frac{0,6-\frac{3}{11}+\frac{3}{13}}{1,4-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(B=\frac{\frac{3}{5}-\frac{3}{11}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(B=\frac{3\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(B=\frac{3}{5}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(B=\frac{3}{5}-\frac{2}{7}=\frac{11}{35}\)
\(A=\frac{7\times9+14\times27+21\times36}{21\times27+42\times81+63\times108}=\frac{7\times9+7\times2\times9\times3+7\times3\times9\times4}{21\times27+21\times2\times27\times3+21\times3\times27\times4}=\frac{7\times9\times\left(1+2\times3+3\times4\right)}{21\times27\times\left(1+2\times3\times3\times4\right)}=\frac{7\times3\times3}{7\times3\times3\times9}=\frac{1}{9}\)
\(A=\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{6}-0,875+0,7}\)
\(=\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{2}{7}\)
TRên tử : đưa 2 ra ngoài
Dưới mẫu: đưa 7 ra
=> A = 2/7