Violympic toán 6
Các câu hỏi tương tự
1,
a,tính:\(\dfrac{\dfrac{7}{2012}+\dfrac{7}{9}-\dfrac{1}{4}}{\dfrac{5}{9}-\dfrac{1}{2012}-\dfrac{1}{2}}\)
b,so sánh:A=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2010};B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{17}\)
Câu 3:
a)\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
b)\(\left(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2013}\right).x=\dfrac{2012}{1}+\dfrac{2011}{2}+\dfrac{2010}{3}+.....+\dfrac{2}{2011}+\dfrac{1}{2012}\)
Tính:
a) \(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)b) \(B=\dfrac{1-3}{1\cdot3}+\dfrac{2-4}{2\cdot4}+\dfrac{3-5}{3\cdot5}+\dfrac{4-6}{4\cdot6}+...+\dfrac{2011-2013}{2011\cdot2013}+\dfrac{2012-2014}{2012\cdot2014}+\dfrac{2013-2015}{2013\cdot2015}\)Giúp mình với!
Rút gọn:
a) A= \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}}{\dfrac{2013}{1}+\dfrac{2014}{2}+...+\dfrac{4024}{2012}-2012}\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right).x=\dfrac{2012}{1}+\dfrac{2011}{2}+...\dfrac{1}{2012}\)
Câu 1:
a) Cho S dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+............+dfrac{1}{2^{2012}}+dfrac{1}{2^{2013}}. Chứng tỏ S1
b) So Sánh: Adfrac{2011^{2012}+1}{2011^{2013}+1} với Bdfrac{2011^{2013}+1}{2011^{2014}+1}
c) So Sánh: C3^{210}với D2^{310}
Đọc tiếp
Câu 1:
a) Cho S= \(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+............+\(\dfrac{1}{2^{2012}}+\dfrac{1}{2^{2013}}\). Chứng tỏ S<1
b) So Sánh: A=\(\dfrac{2011^{2012}+1}{2011^{2013}+1}\) với B=\(\dfrac{2011^{2013}+1}{2011^{2014}+1}\)
c) So Sánh: C=\(3^{210}\)với D=\(2^{310}\)
Bài 2 : So sánh
\(A=\dfrac{2008}{2009}+\dfrac{2009}{2010}+\dfrac{2010}{2011}vàB=\dfrac{2008+2009+2010}{2009+2010+2011}\)
THỰC HIỆN PHÉP TÍNH SAU
P=50% .\(\dfrac{4}{3}.10.\dfrac{7}{35}.0,75\)
Q=(\(\dfrac{1}{3}-25\%-\dfrac{1}{12}\)).\(\left(\dfrac{-789}{2011}+\dfrac{5}{2012}-17\right)\)
Không quy đồng phân số, hãy so sánh: A = \(\dfrac{-9}{10^{2010}}+\dfrac{-19}{10^{2011}}\)và B = \(\dfrac{-9}{10^{2011}}+\dfrac{-19}{10^{2010}}\)