tính tổng:\(\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+......+\dfrac{2018}{2017.2018}\)
Tính tổng:
S = \(\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+...+\dfrac{2018}{2017.2018}\)
\(S=\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+......+\dfrac{2018}{2017.2018}\)
\(S=\dfrac{2018}{2.3}+\dfrac{2018}{3.4}+\dfrac{2018}{4.5}+.......+\dfrac{2018}{2017.2018}\)
\(S=2018\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+......+\dfrac{1}{2017.2018}\right)\)
\(S=2018\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+.....+\dfrac{1}{2017}-\dfrac{1}{2018}\right)\)
\(S=2018\left(\dfrac{1}{2}-\dfrac{1}{2018}\right)\)
\(S=2018\cdot\dfrac{504}{1009}\)
\(S=2.504\)
\(S=1008\)
S = \(\dfrac{2018}{6}+\dfrac{2018}{12}+\dfrac{2018}{20}+...+\dfrac{2018}{2017.2018
}\)
S = \(\dfrac{2018}{2.3}+\dfrac{2018}{3.4}+\dfrac{2018}{4.5}+...+\dfrac{2018}{2017.2018}\)
S = \(\dfrac{2018}{2}-\dfrac{2018}{3}+\dfrac{2018}{3}-\dfrac{2018}{4}+...+\dfrac{2018}{2017}-\dfrac{2018}{2018}\)
S = \(\dfrac{2018}{2}-\dfrac{2018}{2018}\)
S = 1009 - 1
S = 1008
so sánh
1) A = \(\dfrac{10^{11}-1}{10^{12}-1}\) và B =\(\dfrac{10^{10}+1}{10^{11}+1}\)
2) A = \(\dfrac{2018^9+1}{2018^{10}-1}\) và B = \(\dfrac{2018^{19}+1}{2018^{20}+1}\)
3) A = \(\dfrac{2018^{19}+1}{2018^{20}+1}\) và B = \(\dfrac{2018^{20}+1}{2018^{21}+1}\)
Chứng minh: \(A=1.2.3.....2017.2018\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)⋮2019\)
\(A=1.2.3...2018\left[\left(1+\dfrac{1}{2018}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2017}\right)+...+\left(\dfrac{1}{1009}+\dfrac{1}{1010}\right)\right]\)
\(A=1.2.3...2018.2019\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\)
\(\dfrac{A}{2019}=1.2.3...2018\left(\dfrac{1}{1.2018}+\dfrac{1}{2.2017}+...+\dfrac{1}{1009.1010}\right)\).
Rõ ràng tích 1 . 2 ... 2018 chia hết cho các tích 1 . 2018; 2 . 2017; ...; 1009 . 1010; do đó \(\dfrac{A}{2019}\) là số tự nhiên.
Vậy A chia hết cho 2019.
Với x\(\ne-1\) \(\left(\dfrac{x^2+2x+2}{x+1}\right)^{2018}=a_0+a_1x+a_2x^2+...+a_kx^{2018}+\dfrac{b_1}{x+1}+\dfrac{b_2}{\left(x+1\right)^2}+...+\dfrac{b_{2018}}{\left(x+1\right)^{2018}}.\). Tính: S=\(\sum\limits^{2018}_{k=1}bx\)
Đề bài: So sánh
1, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}với\) 3
2, \(\dfrac{2017}{2018}+\dfrac{2018}{2019}với\dfrac{2017+2018}{2018+2019}\)
Ta có : \(\dfrac{2017+2018}{2018+2019}=\dfrac{2017}{2018+2019}+\dfrac{2018}{2018+2019}\)
Rõ ràng ta thấy : \(\dfrac{2017}{2018}>\dfrac{2017}{2018+2019}\) (1)
\(\dfrac{2018}{2019}>\dfrac{2018}{2018+2019}\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{2017}{2018}+\dfrac{2018}{2019}>\dfrac{2017+2018}{2018+2019}\)
Vậy ......................
~ Học tốt ~
Ta có : \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}=\left(1-\dfrac{1}{2018}\right)+\left(1-\dfrac{1}{2019}\right)+\left(1-\dfrac{1}{2020}\right)\)\(=\left(1+1+1\right)-\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)\)
\(=3+\left(\dfrac{1}{2018}+\dfrac{1}{2019}+\dfrac{1}{2020}\right)< 3\)
Vậy \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
Cho dãy số thực: \(a_1,a_2,a_3,............a_{2018}\) thỏa mãn: \(a^1_1+a^2_2+a^3_3+...............+a^{2018}_{2018}=1009\). CM: \(\left(\dfrac{a_1}{1}+\dfrac{a_2}{2}+\dfrac{a_3}{3}+.........+\dfrac{a_{2018}}{2018}\right)^2< 2018\)
Cho dãy số thực: \(a_1,a_2,a_3,............a_{2018}\) thỏa mãn: \(a^1_1+a^2_2+a^3_3+...............+a^{2018}_{2018}=1009\). CM: \(\left(\dfrac{a_1}{1}+\dfrac{a_2}{2}+\dfrac{a_3}{3}+.........+\dfrac{a_{2018}}{2018}\right)^2< 2018\)
Cho dãy số thực: \(a_1,a_2,a_3,.............,a_{2018}\) thỏa mãn: \(a^1_1+a^2_2+a^3_3+.................+a_{2018}^{2018}=1009\). Chứng minh: \(\left(\dfrac{a_1}{1}+\dfrac{a_2}{2}+\dfrac{a_3}{3}+.............+\dfrac{a_{2018}}{2018}\right)^2< 2018\)
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\) Chứng minh rằng:
a, \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b \(\dfrac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\dfrac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\) (1)
a) Từ (1) ta có:
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\) (2)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\) (3)
Từ (2) và (3) suy ra \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
b) Từ (1) ta có:
\(\dfrac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\dfrac{b^{2018}.k^{2018}+d^{2018}.k^{2018}}{b^{2018}+d^{2018}}=\dfrac{k^{2018}\left(b^{2018}+d^{2018}\right)}{b^{2018}+d^{2018}}=k^{2018}\) (4)
\(\dfrac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}=\dfrac{\left(bk+dk\right)^{2018}}{\left(b+d\right)^{2018}}=\dfrac{\left[k\left(b+d\right)\right]^{2018}}{\left(b+d\right)^{2018}}=k^{2018}\) (5)
Từ (4) và (5) suy ra \(\dfrac{a^{2018}+c^{2018}}{b^{2018}+d^{2018}}=\dfrac{\left(a+c\right)^{2018}}{\left(b+d\right)^{2018}}\)