\(\frac{1}{4}.\frac{2}{3}-\frac{3}{2}.\frac{1}{6}+\frac{1}{12}\)

\(=\frac{1}{6}+\frac{-1}{4}+\frac{1}{12}\)

\(=\frac{2}{12}+\frac{-3}{12}+\frac{1}{12}=\frac{0}{12}=0\)

~ 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

\(2\frac{1}{3}+2\frac{5}{7}-1\frac{1}{3}\)

\(=\frac{7}{3}+\frac{19}{7}+\frac{-4}{3}\)

\(=\left(\frac{7}{3}+\frac{-4}{3}\right)+\frac{19}{7}\)

\(=1+\frac{19}{7}\)

\(=\frac{26}{7}\)

~ Hok tốt ~

Bạn cần gõ lại đề bằng công thức toán thật đầy đủ (bộ gõ nằm ở biểu tượng $\sum$ ) để được hỗ trợ tốt hơn, cũng để người nhìn đỡ đau mắt. Bài này sẽ bị xóa sau khi bạn có bài post mới!

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 

\(A=...\)

\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=\left[\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right]:\left(1^2+2^2+...+2015^2\right)\)

\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=0:\left(1^2+2^2+...+2015^2\right)=0\)

\(\dfrac{0,4-\dfrac{2}{9}+\dfrac{2}{11}}{1,4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0,25+\dfrac{1}{5}}{1\dfrac{1}{6}-0,875+0,7}\\ =\dfrac{2\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}{7\left(0,2-\dfrac{1}{9}+\dfrac{1}{11}\right)}-\dfrac{2\left(\dfrac{1}{6}-0,125+0,1\right)}{7\left(\dfrac{1}{6}-0,125+0,1\right)}\\ =\dfrac{2}{7}-\dfrac{2}{7}\\ =0\)

minh cũng đang muốn hỏi câu này nè

\(B=2016:\left(\frac{0.4-\frac{2}{9}+\frac{2}{11}}{1.4-\frac{7}{9}+\frac{7}{11}}.\frac{-1\frac{1}{6}+0.875-0.7}{\frac{1}{3}-0.25+\frac{1}{5}}\right)\) 

<=>\(B=2016:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{\frac{-7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right)\) 

<=>\(B=2016:\left(\frac{2.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}{5.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}.\frac{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}-\frac{2}{8}-\frac{2}{10}}\right)\) 

<=>\(B=2016:\left(\frac{2}{5}.\frac{7.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}\right)\) 

<=>\(B=2016:\left(\frac{2}{5}.\frac{7}{2}\right)\) 

<=>\(B=2016:\frac{7}{5}\) 

<=>\(B=2016.\frac{5}{7}\) 

<=>\(B=1440\) 

 Vậy B=1440

k cho mink nha

Có:

\(A=2010\cdot\left(\frac{\frac{1}{6}+0.25-\frac{1}{8}}{1+1\frac{1}{2}-\frac{3}{4}}+\frac{0.4-\frac{2}{9}+\frac{2}{11}}{3-\frac{15}{9}+1\frac{4}{11}}\right)\)

\(=2010\cdot\left(\frac{\frac{1}{6}+\frac{1}{4}-\frac{1}{8}}{\frac{3}{3}+\frac{3}{2}-\frac{3}{4}}+\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{15}{5}-\frac{15}{9}+\frac{15}{11}}\right)\)

\(=2010\cdot\left[\frac{\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right)}{3\cdot\left(\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right)}+\frac{2\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{15\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\right]\)

\(=2010\cdot\left(\frac{\frac{1}{2}}{3}+\frac{2}{15}\right)\)

\(=2010\cdot\left(\frac{\frac{5}{2}}{15}+\frac{2}{15}\right)\\ =2010\cdot\left(\frac{\frac{5}{2}+2}{15}\right)\)

\(=2010\cdot\frac{\frac{9}{2}}{15}\\ =\frac{2010\cdot\frac{9}{2}}{15}\\ =\frac{1005\cdot9}{15}\\ =201\cdot3\\ =603\)