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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
\(A=...\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=\left[\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right]:\left(1^2+2^2+...+2015^2\right)\)
\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=0:\left(1^2+2^2+...+2015^2\right)=0\)
\(B=2016:\left(\frac{0.4-\frac{2}{9}+\frac{2}{11}}{1.4-\frac{7}{9}+\frac{7}{11}}.\frac{-1\frac{1}{6}+0.875-0.7}{\frac{1}{3}-0.25+\frac{1}{5}}\right)\)
<=>\(B=2016:\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}.\frac{\frac{-7}{6}+\frac{7}{8}-\frac{7}{10}}{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}\right)\)
<=>\(B=2016:\left(\frac{2.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}{5.\left(\frac{1}{5}.\frac{1}{9}.\frac{1}{11}\right)}.\frac{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}{\frac{2}{6}-\frac{2}{8}-\frac{2}{10}}\right)\)
<=>\(B=2016:\left(\frac{2}{5}.\frac{7.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}\right)\)
<=>\(B=2016:\left(\frac{2}{5}.\frac{7}{2}\right)\)
<=>\(B=2016:\frac{7}{5}\)
<=>\(B=2016.\frac{5}{7}\)
<=>\(B=1440\)
Vậy B=1440
k cho mink nha
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
cái dấu trước dấu chia mình ko biết cách đóng ngoặc nhọn lại nên mình viết như vậy
Có:
\(A=2010\cdot\left(\frac{\frac{1}{6}+0.25-\frac{1}{8}}{1+1\frac{1}{2}-\frac{3}{4}}+\frac{0.4-\frac{2}{9}+\frac{2}{11}}{3-\frac{15}{9}+1\frac{4}{11}}\right)\)
\(=2010\cdot\left(\frac{\frac{1}{6}+\frac{1}{4}-\frac{1}{8}}{\frac{3}{3}+\frac{3}{2}-\frac{3}{4}}+\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{15}{5}-\frac{15}{9}+\frac{15}{11}}\right)\)
\(=2010\cdot\left[\frac{\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right)}{3\cdot\left(\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right)}+\frac{2\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{15\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\right]\)
\(=2010\cdot\left(\frac{\frac{1}{2}}{3}+\frac{2}{15}\right)\)
\(=2010\cdot\left(\frac{\frac{5}{2}}{15}+\frac{2}{15}\right)\\ =2010\cdot\left(\frac{\frac{5}{2}+2}{15}\right)\)
\(=2010\cdot\frac{\frac{9}{2}}{15}\\ =\frac{2010\cdot\frac{9}{2}}{15}\\ =\frac{1005\cdot9}{15}\\ =201\cdot3\\ =603\)