Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 5
B= \(\dfrac{2015}{2016+2017+2018}\)+\(\dfrac{2016}{2016+2017+2018}\)+\(\dfrac{2017}{2016+2017+2018}\)
Ta có:\(\dfrac{2015}{2016}\)>\(\dfrac{2015}{2016+2017+2018}\),\(\dfrac{2016}{2017}\)>\(\dfrac{2016}{2016+2017+2018}\),\(\dfrac{2017}{2018}\)>\(\dfrac{2017}{2016+2017+2018}\)
⇒A>B
Bài 5:
Ta có:
\(B=\dfrac{2015+2016+2017}{2016+2017+2018}\)
\(B=\dfrac{2015}{2016+2017+2018}+\dfrac{2016}{2016+2017+2018}+\dfrac{2017}{2016+2017+2018}\)
Vì \(\dfrac{2015}{2016}>\dfrac{2015}{2016+2017+2018}\)
\(\dfrac{2016}{2017}>\dfrac{2016}{2016+2017+2018}\)
\(\dfrac{2017}{2018}>\dfrac{2017}{2016+2017+2018}\)
\(\Rightarrow A>B\)
\(A=1\cdot4+2\cdot5+3\cdot6+...+99\cdot102\)
\(=1\cdot\left(2+2\right)+2\cdot\left(2+3\right)+3\cdot\left(2+4\right)+...+99\cdot\left(2+100\right)\)
\(=\left(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\right)+\left(2+4+6+...+198\right)\)
Ta thấy : \(1\cdot2+2\cdot3+3\cdot4+...+99\cdot100\)nhân với 3 được :
\(1\cdot2\cdot3+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+99\cdot100\cdot\left(101-98\right)\)
\(=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+99\cdot100\cdot101-98\cdot99\cdot100\)
\(=99\cdot100\cdot101\)
\(=999900\)
\(\Rightarrow1\cdot2+2\cdot3+3\cdot4+...+99\cdot100=999900:3=333300\)
\(2+4+6+...+198=\left(198-2\right):2+1=99\)( số hạng )
Tổng của \(2+4+6+...+198\)bằng : \(\left(198+2\right)\cdot99:2=9900\)
\(\Rightarrow A=333300+9900=343200\)
Vậy \(A=343200\)
-2017 + 2789 - 2017 + 1789
= -2017+ 2017+ 2789 -1789
= 0 + 100
= 100
Ư(26)={1;2;13;26}
Ư(39)={1;3;13;39}
Ư(64)={1;2;4;8;16;32;64}
Ư(12)={1;2;3;4;12}
Ư(45)={3;5;9;15;45}
\(Ư\left(26\right)=\left\{1;-1;2;-2;13;-13;21;-26\right\}\)
\(Ư\left(39\right)=\left\{1;-1;3;-3;13;-13;39;-39\right\}\)
\(Ư\left(64\right)=\left\{1;-1;2;-2;4;-4;8;-8;16;-16;32;-32;64;-64\right\}\)
\(Ư\left(12\right)=\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)
\(Ư\left(45\right)=\left\{1;-1;3;-3;5;-5;9;-9;15;-15;45;-45\right\}\)
muốn\(\frac{\left(a-b\right)}{a}\)=\(\frac{\left(c-d\right)}{c}\)
thì\(\frac{\left(c-d\right)}{c}\)=n.\(\frac{\left(a-b\right)}{a}\)(n\(\in\)N*)
giả sử n=2 => ax2=c;bx2=d;(c-d),(a-b)\(⋮\)2 (c>d;a>b)