Tính nhanh:
a, A= \(\frac{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{99.1}}\)
b, B=\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}{\frac{99}{1}+\frac{98}{2}+\frac{97}{3}+...+\frac{1}{99}}\)
a) Cho \(A=\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{100^2}\)
C/mR: \(\frac{1}{6}
Tính giá trị của biểu thức \(\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{4026}}{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{4025}}\)
Bài 1: Cho: \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\) và \(B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\) Tìm \(\frac{A}{B}\)
Bài 2: Tính: \(M=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{2014^2}\)
Bài 3: Tìm x;y biết: \(A=\left(x-y\right)^2+\left(2x+3y-10\right)^2-2\)
tìm x,y biết:
a)\(\left(2x-5\right)^{2000}+\left(3y+4\right)^{2002}\le0\)
b)\(x=6:\frac{1}{3}-0,8:\frac{1,5}{\frac{3}{2}.0,4.\frac{50}{1:\frac{1}{2}}}+\frac{1}{4}+\frac{1+\frac{1}{2}.\frac{1}{0,25}}{6-\frac{46}{1+2,2.10}}\)
c)\(y=77^{-1}.7^4.11^2.\left(1.11\right)^4.\left(7^2\right)^{-8}.\left(7^{-8}\right)^{-3}.\frac{1}{\left(-11\right)^{-3}}\)
Tính giá trị của các biểu thức sau:
\(A=\frac{\frac{3}{4}-\frac{3}{26}+\frac{3}{17}+\frac{3}{11}}{\frac{7}{4}-\frac{7}{26}+\frac{7}{1}+\frac{7}{11}}:\frac{\frac{5}{6}+\frac{5}{9}-\frac{5}{12}}{\frac{7}{3}+\frac{14}{9}-\frac{7}{6}}\)
\(B=\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+.........+\frac{1}{34.39}\)
\(C=\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-\frac{1}{97.96}-........-\frac{1}{3.2}-\frac{1}{2.1}\)
tính giá trị của biểu thức
a,\(-\frac{1}{2}+\frac{5}{6}-3,12+5,1\)
b,\(30+2,8:\left(\frac{4}{25}-\frac{9}{5}-\frac{1}{6}\right)\)
c,\(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right)\times230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
Tính:
a.A = \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b. B = \(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c. C = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
Cho \(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}\)
\(B=\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+....+\frac{2}{2010}+\frac{1}{2011}\)
Tính \(\frac{A}{B}\)