Violympic toán 6
Các câu hỏi tương tự
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!
Câu 3:
a)\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
b)\(\left(\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}+\dfrac{1}{2013}\right).x=\dfrac{2012}{1}+\dfrac{2011}{2}+\dfrac{2010}{3}+.....+\dfrac{2}{2011}+\dfrac{1}{2012}\)
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=\(\dfrac{2010}{2011}+\dfrac{2011}{2012}+\dfrac{2012}{2010};B=\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+...+\dfrac{1}{17}\)
\(\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right).x=\dfrac{2012}{1}+\dfrac{2011}{2}+...\dfrac{1}{2012}\)
Rút gọn:
a) A= \(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}}{\dfrac{2013}{1}+\dfrac{2014}{2}+...+\dfrac{4024}{2012}-2012}\)
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=\(2^{310}\)
Các bạn giúp với :Bài 1:a, CMR: A dfrac{3}{1^2.2^2}+dfrac{5}{2^2.3^2}+dfrac{7}{3^2.4^2}+...+dfrac{21}{10^2.11^2} 1b, Cho B dfrac{3}{4}+dfrac{8}{9}+dfrac{15}{16}+dfrac{24}{25}+...+dfrac{2499}{2500}. CMR: B không phải là số nguyên.c, So sánh: C dfrac{2}{2^1}+dfrac{3}{2^2}+dfrac{4}{2^3}+...+dfrac{2021}{2^{2020}} với 3.
Đọc tiếp
Các bạn giúp với :<
Bài 1:
a, CMR: A = \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{21}{10^2.11^2}< 1\)
b, Cho B = \(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+\dfrac{24}{25}+...+\dfrac{2499}{2500}.\) CMR: B không phải là số nguyên.
c, So sánh: C = \(\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2021}{2^{2020}}\) với 3.
Cho A dfrac{1}{2014}+dfrac{2}{2013}+dfrac{3}{2012}+...+dfrac{2013}{2}+2014 B dfrac{1}{2}+dfrac{1}{3}+dfrac{1}{4}+...+dfrac{1}{2015} Tính giá trị dfrac{A}{B}
Đọc tiếp
Cho A = \(\dfrac{1}{2014}\)+\(\dfrac{2}{2013}\)+\(\dfrac{3}{2012}\)+...+\(\dfrac{2013}{2}\)+2014
B = \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{2015}\)
Tính giá trị \(\dfrac{A}{B}\)
CMR : \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Với \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\)