ai giải giúp mình bài này với
\(M=\left(1+\frac{1}{1+2}\right)\left(1+\frac{1}{1+2+3}\right)......\left(1+\frac{1}{1+2+3+...+2012}\right)\)
please help me
Tính ; A =1+\(\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2011}\left(1+2+3+...+2011\right)\)
Bài 1: Tính nhanh
\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{2011}\right)\)
Bài 2:
So sánh\(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}\)với 3
a, Tính : \(\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{10}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
b, Tính : \(P=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}}{\frac{2011}{1}+\frac{2010}{2}+\frac{2009}{3}+...+\frac{1}{2011}}\)
c, Tính : \(\frac{\left(1+2+3+...+99+100\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right)\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right)......\left(1-\frac{1}{1+2+3+4+....+2011}\right)\)
Bài 1: Tính nhanh
\(\left(1-\frac{1}{7}\right)\times\left(1-\frac{1}{8}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{2011}\right)\)
Bài 2:
So sánh \(\frac{2011}{2012}+\frac{2012}{20013}+\frac{2013}{2011}\)với 3
Thực hiện phép tính
a. \(25^{10}-\left(\frac{1}{5}\right)^{20}+\left(\frac{-3}{4}\right)^8-2011^0\)
b. \(\left(\frac{1}{2}-\frac{2}{3}\right)-\left(\frac{5}{3}-\frac{3}{2}\right)-\left(\frac{7}{3}-\frac{5}{2}\right)\)
Các bạn giúp mk với, mk đang cần gấp
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 = \(\frac{\left(1+2+3+...+99+100\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{7}-\frac{1}{9}\right).\left(63.1,2-21.3,6\right)}{1-2+3-4+...+99-100}\)
A = \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot.....\cdot\left(1+\frac{1}{2011\cdot2013}\right)\)