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Sửa đề: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
Ta có: \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(=\frac{-3\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(=\frac{-3}{5}+\frac{3}{5}=0\)
\(A=\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
\(A=\left(\frac{\frac{3}{2}+1-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{1}{2}-\frac{5}{11}-\frac{5}{12}}\right):\frac{378}{401}+115\)
\(A=\left(\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{-3.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5.\left(\frac{-1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}\right).\frac{401}{378}+115\)
\(A=\left(\frac{3}{5}+\frac{-3}{5}\right).\frac{401}{378}+115\)
\(A=0.\frac{401}{378}+115=115\)
A = \(\left(\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}+\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}+\frac{\frac{3.125}{100}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{-\frac{5.125}{100}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{3\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}+\frac{3\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{-5\left(\frac{125}{100}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}\right):\frac{1890}{2005}+115\)
= \(\left(\frac{3}{5}+-\frac{3}{5}\right):\frac{1890}{2005}+115\)
= 115
\(B=\frac{2001}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{2}{2010}+\frac{1}{2001}\)
\(B=\left(2011-1-...-1\right)+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2011}+\frac{2012}{2012}\)
\(B=2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
\(\Rightarrow\)\(\frac{B}{A}=\frac{2012\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}}=2012\)
Vậy \(\frac{B}{A}=2012\)
Chúc bạn học tốt ~
Đặt: \(L=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}\)
\(L=1+\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{1}{2011}+1\right)\)
\(L=\frac{2012}{2012}+\frac{2012}{2}+\frac{2012}{3}+..+\frac{2012}{2011}\)
\(L=2012\left(\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2011}+\frac{1}{2012}\right)\)
Hay: \(P=\frac{1}{2012}\)
có ai kb với mik ko