\(\frac{-11}{12}.x+0.25=\frac{5}{6}\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{10\frac{1}{3}\left(26\frac{1}{3}-\frac{176}{7}\right)-\frac{12}{11}\left(\frac{10}{3}-1,75\right)}{\frac{5}{\left(91-0,25\right).\frac{60}{11}-1}}\)
\(\Leftrightarrow\left(\frac{31}{3}.\frac{25}{21}-\frac{12}{11}.\frac{19}{12}\right):\left(5:495-1\right)\)
\(\Leftrightarrow\left(\frac{775}{63}-\frac{19}{11}\right):\left(-\frac{98}{99}\right)\)
\(\Leftrightarrow x=-\frac{3664}{343}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2016:\left(\frac{0.4-\frac{2}{9}+\frac{2}{11}}{1.4-\frac{7}{9}+\frac{7}{11}}.\frac{-1\frac{1}{6}+0.875-0.7}{\frac{1}{3}-0.25+\frac{1}{5}}\right)\)
<=>\(B=2016:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{\frac{-7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right)\)
<=>\(B=2016:\left(\frac{2.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}{5.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}.\frac{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}-\frac{2}{8}-\frac{2}{10}}\right)\)
<=>\(B=2016:\left(\frac{2}{5}.\frac{7.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}\right)\)
<=>\(B=2016:\left(\frac{2}{5}.\frac{7}{2}\right)\)
<=>\(B=2016:\frac{7}{5}\)
<=>\(B=2016.\frac{5}{7}\)
<=>\(B=1440\)
Vậy B=1440
k cho mink nha
![](https://rs.olm.vn/images/avt/0.png?1311)
Trả lời
\(A=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right).\)
\(A=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+3^2+...+2015^2\right)\)
\(A=0:\left(1^2+2^2+3^2+.....+2015^2\right)\)
A=0
Study well
\(A=...\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=\left[\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right]:\left(1^2+2^2+...+2015^2\right)\)
\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=0:\left(1^2+2^2+...+2015^2\right)=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
cái dấu trước dấu chia mình ko biết cách đóng ngoặc nhọn lại nên mình viết như vậy
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{4}.\frac{2}{3}-\frac{3}{2}.\frac{1}{6}+\frac{1}{12}\)
\(=\frac{1}{6}+\frac{-1}{4}+\frac{1}{12}\)
\(=\frac{2}{12}+\frac{-3}{12}+\frac{1}{12}=\frac{0}{12}=0\)
~ Hok tốt ~
A, \(=\frac{1}{6}-\frac{5}{3}+\frac{1}{12}\)
\(=\frac{2}{12}-\frac{20}{12}+\frac{1}{12}\)
\(=\frac{2-20+1}{12}=-\frac{17}{12}\)
Hok tốt
\(\frac{-11}{12}\) .x+0.25 = \(\frac{5}{6}\)
\(\frac{-11}{12}\) .x+ \(\frac{1}{4}\)= \(\frac{5}{6}\)
\(\frac{-11}{12}\).x =\(\frac{5}{6}\)- \(\frac{1}{4}\)
\(\frac{-11}{12}\).x = \(\frac{7}{10}\)
x = \(\frac{7}{10}\)- \(\frac{-11}{12}\)
x = \(\frac{47}{60}\)
k cho mik nha,chúc bn học tốt
\(\frac{-11}{12}\cdot x+0,25=\frac{5}{6}\)
\(\Rightarrow\frac{-11}{12}\cdot x+\frac{1}{4}=\frac{5}{6}\)
\(\Rightarrow\frac{-11}{12}\cdot x=\frac{5}{6}-\frac{1}{4}\)
\(\Rightarrow\frac{-11}{12}\cdot x=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}:\frac{-11}{12}\)
\(\Rightarrow x=\frac{7}{-11}\)
vay___