Violympic toán 9
Các câu hỏi tương tự
cmr:\(a=1.2.3...2003.2004.\left(1+\dfrac{1}{2}+\dfrac{1}{2}+...+\dfrac{1}{2003}+\dfrac{1}{2004}\right)\)chia hết cho 2005
Chứng minh: \(A=1.2.3.....2017.2018\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)⋮2019\)
Cho a,b,c \(\ge1.CMR:a\left(b+c\right)+b\left(c+a\right)+c\left(a+b\right)+2\left(\dfrac{1}{1+a^2}+\dfrac{1}{1+b^2}+\dfrac{1}{1+c^2}\right)\ge9\)
Tìm số tự nhiên n sao cho:
\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{n\left(n+1\right)\left(n+2\right)}=\dfrac{637}{2550}\)
Chứng minh: Adfrac{2^3+1}{2^3-1}.dfrac{3^3+1}{3^3-1}.dfrac{4^3+1}{4^3-1}....dfrac{9^3+1}{9^3-1} dfrac{3}{2}
Bdfrac{1}{2!}+dfrac{1}{3!}+dfrac{1}{4!}+....+dfrac{1}{n!} 1
Cdfrac{1}{2!}+dfrac{2}{3!}+dfrac{3}{4!}+....+dfrac{n-1}{n!} 1
Dleft(1-dfrac{2}{6}right)left(1-dfrac{2}{12}right)left(1-dfrac{2}{20}right)....left(1-dfrac{2}{nleft(n+1right)}right)dfrac{1}{3}
Đọc tiếp
Chứng minh: \(A=\dfrac{2^3+1}{2^3-1}.\dfrac{3^3+1}{3^3-1}.\dfrac{4^3+1}{4^3-1}....\dfrac{9^3+1}{9^3-1}< \dfrac{3}{2}\)
\(B=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+....+\dfrac{1}{n!}< 1\)
\(C=\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+....+\dfrac{n-1}{n!}< 1\)
D=\(\left(1-\dfrac{2}{6}\right)\left(1-\dfrac{2}{12}\right)\left(1-\dfrac{2}{20}\right)....\left(1-\dfrac{2}{n\left(n+1\right)}\right)>\dfrac{1}{3}\)
Tính các tổng sau:
Adfrac{1}{2.5}+dfrac{1}{5.8}+dfrac{1}{8.11}+.....+dfrac{1}{left(3n-1right)left(3n+2right)}
Bdfrac{1}{1.3.5}+dfrac{1}{3.5.7}+dfrac{1}{5.7.9}+....+dfrac{1}{left(2n-1right)left(2n+1right)left(2n+3right)}
Csqrt{1+dfrac{1}{1^2}+dfrac{1}{2^2}}+sqrt{1+dfrac{1}{2^2}+dfrac{1}{3^2}}+sqrt{1+dfrac{1}{3^2}+dfrac{1}{4^2}}+....+sqrt{1+dfrac{1}{2018^2}+dfrac{1}{2019^2}}
Đọc tiếp
Tính các tổng sau:
A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{\left(3n-1\right)\left(3n+2\right)}\)
B=\(\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{\left(2n-1\right)\left(2n+1\right)\left(2n+3\right)}\)
C=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+....+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
Tính các tích sau:
P_1 left(1+dfrac{2}{4}right)left(1+dfrac{2}{10}right)left(1+dfrac{2}{18}right)....left(1+dfrac{2}{n^2+3n}right)
P_2 left(1+dfrac{1}{3}right)left(1+dfrac{1}{8}right)left(1+dfrac{1}{15}right)....left(1+dfrac{2}{n^2+2n}right)
P_3 left(1-dfrac{1}{1+2}right)left(1-dfrac{1}{1+2+3}right)left(1-dfrac{1}{1+2+3+4}right).....left(1-dfrac{1}{1+2+3+4+...+n}right)
P_4 dfrac{2^4+4}{4^4+4}.dfrac{6^4+4}{8^4+4}.dfrac{8^4+4}{10^4+4}....dfrac{18^4+4}{20^4+4}
Đọc tiếp
Tính các tích sau:
P\(_1\) =\(\left(1+\dfrac{2}{4}\right)\left(1+\dfrac{2}{10}\right)\left(1+\dfrac{2}{18}\right)....\left(1+\dfrac{2}{n^2+3n}\right)\)
P\(_2\) =\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{2}{n^2+2n}\right)\)
P\(_3\) = \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right).....\left(1-\dfrac{1}{1+2+3+4+...+n}\right)\)
P\(_4\) = \(\dfrac{2^4+4}{4^4+4}.\dfrac{6^4+4}{8^4+4}.\dfrac{8^4+4}{10^4+4}....\dfrac{18^4+4}{20^4+4}\)
cho \(7\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)=6\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)+2017\)
tìm max \(\dfrac{1}{\sqrt{3\left(2a^2+b^2\right)}}+\dfrac{1}{\sqrt{3\left(2b^2+c^2\right)}}+\dfrac{1}{\sqrt{3\left(2c^2+a^2\right)}}\)
Rút gọn: \(A=\dfrac{\left(1+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)\left(5^4+\dfrac{1}{4}\right)...\left(51^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)\left(6^4+\dfrac{1}{4}\right)...\left(52^4+\dfrac{1}{4}\right)}\)