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Đặt \(\frac{2016}{2017}\)+\(\frac{2017}{2018}\)+\(\frac{2018}{2019}\)+\(\frac{2019}{2016}\) là A
A=1-\(\frac{1}{2017}\)+1-\(\frac{1}{2018}\)+1-\(\frac{1}{2019}\)+1+\(\frac{3}{2016}\)
A=4-(\(\frac{1}{2017}\)+\(\frac{1}{2018}\)+\(\frac{1}{2019}\)-\(\frac{3}{2016}\)) Do \(\frac{1}{2017}\)+\(\frac{1}{2018}\)+\(\frac{1}{2019}\)-\(\frac{3}{2016}\)<0 =>A>4
Giải:
Ta có:
\(P=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
và \(Q=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Vì \(\left\{{}\begin{matrix}\dfrac{2016}{2017}=\dfrac{2016}{2017}\\\dfrac{2017}{2018}=\dfrac{2017}{2018}\\\dfrac{2018}{2019}=\dfrac{2018}{2019}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}=\dfrac{2016}{2017}+\dfrac{2017}{2018}+\dfrac{2018}{2019}\)
Hay \(P=Q\)
Vậy ...
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 \(\dfrac{2017}{2018}+\dfrac{2018}{2019}+\dfrac{2019}{2020}< 3\)
\(A=\dfrac{\dfrac{1}{2017}+\dfrac{2}{2016}+\dfrac{3}{2015}+...+\dfrac{2016}{2}+\dfrac{2017}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\left(\dfrac{1}{2017}+1\right)+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{3}{2015}+1\right)+...+\left(\dfrac{2016}{2}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{\dfrac{2018}{2017}+\dfrac{2018}{2016}+\dfrac{2018}{2015}+...+\dfrac{2018}{2}+\dfrac{2018}{2018}}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}\)
\(A=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2016}+\dfrac{1}{2017}+\dfrac{1}{2018}}=2018\)
\(\dfrac{2016+x}{2017+x}\)=\(\dfrac{2018}{2017}\)
1-\(\dfrac{2016+x}{2017+x}=1-\dfrac{2018}{2017}\)
\(\dfrac{2017+x}{2017+x}-\dfrac{2016+x}{2017+x}=\dfrac{2017}{2017}-\dfrac{2018}{2017}\)
\(\dfrac{\left(2017+x\right)-\left(2016+x\right)}{2017+x}\)=\(\dfrac{2017-2018}{2017}\)
\(\dfrac{2017+x-2016-x}{2017+x}\) = \(\dfrac{-1}{2017}\)
\(\dfrac{\left(2017-2016\right)+\left(x-x\right)}{2017+x}\)= \(\dfrac{1}{-2017}\)
\(\dfrac{1}{2017+x}\) = \(\dfrac{1}{-2017}\)
2017+x = -2017
x = (-2017)-2017
x = -4034
Vậy x = -4034
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\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+\dfrac{2017}{4}+...+\dfrac{2017}{2018}}{\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}}\)
Đặt \(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\) là B
\(B=\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{1}{2017}\\ =\dfrac{2017}{1}+1+\dfrac{2016}{2}+1+...+\dfrac{1}{2017}+1-2017\\ =\dfrac{2018}{1}+\dfrac{2018}{2}+...+\dfrac{2018}{2017}-2017\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\left(2018-2017\right)\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+1\\ =\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\\ =2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)\)
\(A=\dfrac{\dfrac{2017}{2}+\dfrac{2017}{3}+...+\dfrac{2017}{2018}}{2018\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}{2018.\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2018}\right)}\\ =\dfrac{2017}{2018}\)
Tìm x \(\in\) Z biết:
1) \(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ...
\(\dfrac{1}{3}+\dfrac{1}{6}+....+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2017}{2019}\\ \Rightarrow2.\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{2017}{2019}\\ \Rightarrow\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{2017}{4038}\\ \Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2019}\\ \Rightarrow x=2018\)
\(\dfrac{x-1}{2019}+\dfrac{x-2}{2018}=\dfrac{x-3}{2017}+\dfrac{x-4}{2016}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2019}-1\right)+\left(\dfrac{x-2}{2018}-1\right)=\left(\dfrac{x-3}{2017}-1\right)+\left(\dfrac{x-4}{2016}-1\right)\)
\(\Leftrightarrow\dfrac{x-2020}{2019}+\dfrac{x-2020}{2018}-\dfrac{x-2020}{2017}-\dfrac{x-2010}{2016}=0\)
\(\Leftrightarrow\left(x-2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Rightarrow x-2020=0\Leftrightarrow x=2020\)
vậy.......