Violympic toán 9
Các câu hỏi tương tự
CMR : \(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{2018}}>2\left(\sqrt{2018}-1\right)\)
Tính giá trị của biểu thức: \(A=\sqrt{1+2017^2+\dfrac{2017^2}{2018^2}}+\dfrac{2017}{2018}\)
So sánh :
A=\(\sqrt{2019^2-1}-\sqrt{2018^2-1}\) và B=\(\dfrac{2.2019}{\sqrt{2019^2-1+\sqrt{2018^2-1}}}\)
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=\(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+\sqrt{1+\dfrac{1}{3^2}+\dfrac{1}{4^2}}+....+\sqrt{1+\dfrac{1}{2018^2}+\dfrac{1}{2019^2}}\)
H24
21 tháng 2 2019 lúc 21:13
CMR: \(\sqrt{1+\dfrac{1}{1^2}+\dfrac{1}{2^2}}+\sqrt{1+\dfrac{1}{2^2}+\dfrac{1}{3^2}}+...+\sqrt{1+\dfrac{1}{2017^2}+\dfrac{1}{2018^2}}< 2018\)
Tính: \(\dfrac{1}{\sqrt{1}}+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}\left(n\in N,n>0\right)\)
Help Nguyễn TrươngNguyễn Việt LâmKhôi Bùi Akai HarumaDƯƠNG PHAN KHÁNH DƯƠNG
TÍNH \(A=\dfrac{1}{\sqrt{1^3}}+\dfrac{1}{\sqrt{1^3+2^3}}+....+\dfrac{1}{\sqrt{1^3+2^3+...+2018^3}}\)
Tìm STN n biết
\(\dfrac{1}{\sqrt{1^3+2^3}}+\dfrac{1}{\sqrt{1^3+2^3+3^3}}+....+\dfrac{1}{\sqrt{1^3+2^3+3^3+...+n^3}}=\dfrac{2017}{2019}\)
So sánh x và y trong các TH sau: \(x=\dfrac{2017}{\sqrt{2018}}+\dfrac{2018}{\sqrt{2017}};y=\sqrt{2017}+\sqrt{2018}\)
1) Chứng minh rằng: 1+dfrac{1}{2sqrt{2}}+dfrac{1}{3sqrt{3}}+...+dfrac{1}{nsqrt{n}} 2sqrt{2}left(nin Nright)
2) Chứng minh rằng: dfrac{2}{3}+sqrt{n+1} 1+sqrt{2}+sqrt{3}+...+sqrt{n} dfrac{2}{3}left(n+1right)sqrt{n}
3) 2sqrt{n}-3 dfrac{1}{sqrt{2}}+dfrac{1}{sqrt{3}}+...+dfrac{1}{sqrt{n}} 2sqrt{n}-2
4) dfrac{sqrt{2}-sqrt{1}}{2+1}+dfrac{sqrt{3}-sqrt{2}}{3+2}+...+dfrac{sqrt{n+1}-sqrt{n}}{n+1+n} dfrac{1}{2}left(1-dfrac{1}{sqrt{n+1}}right)
Đọc tiếp
1) Chứng minh rằng: \(1+\dfrac{1}{2\sqrt{2}}+\dfrac{1}{3\sqrt{3}}+...+\dfrac{1}{n\sqrt{n}}< 2\sqrt{2}\left(n\in N\right)\)
2) Chứng minh rằng: \(\dfrac{2}{3}+\sqrt{n+1}< 1+\sqrt{2}+\sqrt{3}+...+\sqrt{n}< \dfrac{2}{3}\left(n+1\right)\sqrt{n}\)
3) \(2\sqrt{n}-3< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}< 2\sqrt{n}-2\)
4) \(\dfrac{\sqrt{2}-\sqrt{1}}{2+1}+\dfrac{\sqrt{3}-\sqrt{2}}{3+2}+...+\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1+n}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{n+1}}\right)\)