1. So sánh
A = \(\dfrac{2015}{2018^3}\) + \(\dfrac{2017}{2018^4}\) và B = \(\dfrac{2017}{2018^3}\) + \(\dfrac{2015}{2018^4}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
18 tháng 3 2018
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
KC
3
8 tháng 3 2018
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
<=>\(\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{2017}-1+\dfrac{x+4}{2018}-1\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
<=>\(\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
<=>\(\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
vì 1/2015+1/2016-1/2017-1/2018 khác 0
=>x-2014=0<=>x=2014
vậy.....................
chúc bạn học totts ^^
ND
8 tháng 3 2018
\(\dfrac{x+1}{2015}+\dfrac{x+2}{2016}=\dfrac{x+3}{2017}+\dfrac{x+4}{2018}\)
\(\Leftrightarrow\dfrac{x+1}{2015}-1+\dfrac{x+2}{2016}-1=\dfrac{x+3}{x017}-1+\dfrac{x+4}{2018}-1\)
\(\Leftrightarrow\dfrac{x+1-2015}{2015}+\dfrac{x+2-2016}{2016}=\dfrac{x+3-2017}{2017}+\dfrac{x+4-2018}{2018}\)\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}=\dfrac{x-2014}{2017}+\dfrac{x-2014}{2018}\)
\(\Leftrightarrow\dfrac{x-2014}{2015}+\dfrac{x-2014}{2016}-\dfrac{x-2014}{2017}-\dfrac{x-2014}{2018}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Vì: \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Rightarrow x-2014=0\)
\(\Rightarrow x=2014\)
Vậy........
LG
6 tháng 8 2017
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 ~
LG
6 tháng 8 2017
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 \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
1
1
13 tháng 12 2018
\(\dfrac{x+4}{2015}+\dfrac{x+3}{2016}=\dfrac{x+2}{2017}+\dfrac{x+1}{2018}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}=\dfrac{x+2019}{2017}+\dfrac{x+2019}{2018}\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}-\dfrac{x+2019}{2017}-\dfrac{x+2019}{2018}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
Vậy...
TT
1
S
18 tháng 10 2017
\(\dfrac{x-2}{2018}=\dfrac{x-3}{2017}=\dfrac{x-4}{2016}=\dfrac{x-5}{2015}\)
\(\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=\dfrac{x-4}{2016}+\dfrac{x-5}{2015}\)
\(\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=\left(\dfrac{x-4}{2016}-1\right)+\left(\dfrac{x-5}{2015}-1\right)\)
\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=\dfrac{x-2020}{2016}+\dfrac{x-2020}{2015}\)
\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}-\dfrac{x-2020}{2016}-\dfrac{x-2020}{2015}=0\)
\(\left(x-2020\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)
\(\dfrac{1}{2018};\dfrac{1}{2017};\dfrac{1}{2016};\dfrac{1}{2015}>0\)
Nên \(x-2020=0\)
\(x=0+2020\)
\(x=2020\)
Vậy x bằng 2020
TT
1
HH
24 tháng 6 2018
Giải:
\(\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)
\(\Leftrightarrow2+\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=2+\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)
\(\Leftrightarrow\dfrac{x+2015}{5}+1+\dfrac{x+2016}{4}+1=\dfrac{x+2017}{3}+1+\dfrac{x+2018}{2}+1\)
\(\Leftrightarrow\dfrac{x+2015+5}{5}+\dfrac{x+2016+4}{4}=\dfrac{x+2017+3}{3}+\dfrac{x+2018+2}{2}\)
\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}=\dfrac{x+2020}{3}+\dfrac{x+2020}{2}\)
\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}-\dfrac{x+2020}{3}-\dfrac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ...
18 tháng 4 2017
a, Ta có: \(\dfrac{2016}{2017+2018}< \dfrac{2016}{2017}\)
\(\dfrac{2017}{2017+2018}< \dfrac{2017}{2018}\)
\(\Rightarrow A=\dfrac{2016+2017}{2017+2018}< B=\dfrac{2016}{2017}+\dfrac{2017}{2018}\)
Vậy A < B
b, Ta có: \(\dfrac{2017}{2016+2017}< \dfrac{2017}{2016}\)
\(\dfrac{2018}{2016+2017}< \dfrac{2018}{2017}\)
\(\Rightarrow M=\dfrac{2017+2018}{2016+2017}< N=\dfrac{2017}{2016}+\dfrac{2018}{2017}\)
Vậy M < N
26 tháng 3 2017
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(B=\dfrac{2\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(B=\dfrac{2}{3}:\dfrac{4}{5}\) ( Do \(\left\{{}\begin{matrix}1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\ne0\\1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\ne0\end{matrix}\right.\))
\(B=\dfrac{2}{3}\cdot\dfrac{5}{4}=\dfrac{2\cdot5}{3\cdot4}=\dfrac{5}{6}\)
26 tháng 3 2017
\(B=\dfrac{2-\dfrac{2}{19}+\dfrac{2}{43}-\dfrac{2}{2017}}{3-\dfrac{3}{19}+\dfrac{3}{43}-\dfrac{3}{2017}}:\dfrac{4-\dfrac{4}{29}+\dfrac{4}{41}-\dfrac{4}{2018}}{5-\dfrac{5}{29}+\dfrac{5}{41}-\dfrac{5}{2018}}\)
\(\Rightarrow\)\(B=\dfrac{2-\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}{3\left(1-\dfrac{1}{19}+\dfrac{1}{43}-\dfrac{1}{2017}\right)}:\dfrac{4\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}{5\left(1-\dfrac{1}{29}+\dfrac{1}{41}-\dfrac{1}{2018}\right)}\)
\(\Rightarrow B=\dfrac{2}{3}:\dfrac{4}{5}=\dfrac{10}{12}=\dfrac{5}{6}\)
8 tháng 7 2017
Các câu dễ bạn tự làm nha:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(A< \dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\Rightarrow A< \dfrac{2017^{2017}+2017}{2017^{2018}+2017}\Rightarrow A< \dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}\Rightarrow A< \dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)\(A< B\)
