Câu 1: Rút gọn: \(A=\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right):\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)\)
Câu 2: Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\). Tính \(\left(S-P\right)^{2013}\)
Cho \(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\) và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\). Tính \(\left(S-P\right)^{2013}\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
Và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
Tính \(\left(S-P\right)^{2013}\)
Cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
P=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
Tinh \(\left(S-P\right)^{2013}\)
Cho S=\(1-\frac{1}{2}+\frac{1}{3}\)-\(\frac{1}{4}\)+...........+\(\frac{1}{2011}\)-\(\frac{1}{2012}\)+\(\frac{1}{2013}\)và P=\(\frac{1}{1007}\)+\(\frac{1}{1008}\)+.....+\(\frac{1}{2012}\)+\(\frac{1}{2013}\)
Tính \(\left(S-P\right)^{2013}\)
cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
va P=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
tinh (S-P)2013
s=1-\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
và p=\(\frac{1}{1007}+\frac{1}{1008}+....+\frac{1}{2012}+\frac{1}{2013}\)
tính(S-P)^2013
đề thi chọn học sinh giỏi môn toán, lớp 7, tỉnh bắc giang năm học 2012-2013
Cho \(S=1-\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\) và \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
Tính ( S - P )2013
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
cho S=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)và P=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
Tính (S-P)213
