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\(\Leftrightarrow\left(\dfrac{x+1}{2019}+1\right)+\left(\dfrac{x+2}{2018}+1\right)=\left(\dfrac{x+3}{2017}+1\right)+\left(\dfrac{x+4}{2016}+1\right)\)
\(\Leftrightarrow\dfrac{x+2020}{2019}+\dfrac{x+2020}{2018}-\dfrac{x+2020}{2017}-\dfrac{x+2020}{2016}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\right)=0\)
\(\Leftrightarrow x=-2020\)(do \(\dfrac{1}{2019}+\dfrac{1}{2018}-\dfrac{1}{2017}-\dfrac{1}{2016}\ne0\))
a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)
=1-2/4=1/2
b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)
\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)
c: x-y=0 nên x=y
\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)
=2019
\(\Leftrightarrow\dfrac{x-2}{2020}-1+\dfrac{x-3}{2019}-1=\dfrac{x-2019}{3}-1+\dfrac{x-2020}{2}-1\)
=>x-2022=0
hay x=2022
a)
`(2x-1)(x+2/3)=0`
\(< =>\left[{}\begin{matrix}2x-1=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2019}+\dfrac{x+3}{2020}=\dfrac{x+2}{2021}+\dfrac{x+1}{2022}\)
\(< =>\dfrac{x+4}{2019}+1+\dfrac{x+3}{2020}+1=\dfrac{x+2}{2021}+1+\dfrac{x+1}{2022}+1\)
\(< =>\dfrac{x+2023}{2019}+\dfrac{x+2023}{2020}=\dfrac{x+2023}{2021}+\dfrac{x+2023}{2022}\)
\(< =>\left(x+2023\right)\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\right)=0\)
\(< =>x+2023=0\left(\dfrac{1}{2019}+\dfrac{1}{2020}-\dfrac{1}{2021}-\dfrac{1}{2022}\ne0\right)\\ < =>x=-2023\)
\(\dfrac{3-x}{2016}-1=\dfrac{2-x}{2017}+\dfrac{1-x}{2018}\)
\(\Leftrightarrow\dfrac{3-x}{2016}+1=\dfrac{2-x}{2017}+1+\dfrac{1-x}{2018}+1\)
\(\Leftrightarrow\dfrac{2019-x}{2016}=\dfrac{2019-x}{2017}+\dfrac{2019-x}{2018}\)
\(\left(2019-x\right)\left(\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
2019 -x =0 ; x =2019
\(\dfrac{x+4}{2015}+\dfrac{x+3}{2016}=\dfrac{x+2}{2017}+\dfrac{x+1}{2018}\)
\(\Leftrightarrow\left(\dfrac{x+4}{2015}+1\right)+\left(\dfrac{x+3}{2016}+1\right)=\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+1}{2018}+1\right)\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}=\dfrac{x+2019}{2017}+\dfrac{x+2019}{2018}\)
\(\Leftrightarrow\dfrac{x+2019}{2015}+\dfrac{x+2019}{2016}-\dfrac{x+2019}{2017}-\dfrac{x+2019}{2018}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\right)=0\)
Mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2017}-\dfrac{1}{2018}\ne0\)
\(\Leftrightarrow x+2019=0\)
\(\Leftrightarrow x=-2019\)
Vậy...
\(\dfrac{2017}{1}+\dfrac{2016}{2}+...+\dfrac{2}{2016}+\dfrac{1}{2017}\)
\(=\left(\dfrac{2016}{2}+1\right)+\left(\dfrac{2015}{3}+1\right)+...+\left(\dfrac{2}{2016}+1\right)+\left(\dfrac{1}{2017}+1\right)+1\)
\(=\dfrac{2018}{2}+\dfrac{2018}{3}+...+\dfrac{2018}{2017}+\dfrac{2018}{2018}\)
\(=2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)\)
Theo đề, ta có: \(x=\dfrac{2018\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2018}}=2018\)
\(\dfrac{1}{2019}:2017.x=-\dfrac{1}{2017}\)
\(\dfrac{1}{2019.2017}x=-\dfrac{1}{2017}\)
x=\(-\dfrac{1}{2017}:\dfrac{1}{2019.2017}\)=-2019
Vậy x=-2019
2017 . x = \(\dfrac{1}{2019}:\left(\dfrac{-1}{2017}\right)\)
2017 . x = \(\dfrac{1}{2019}.\left(-2017\right)\)
2017 . x = \(-\dfrac{2017}{2019}\)
x = \(-\dfrac{2017}{2019}:2017\)
x = \(-\dfrac{2017}{2019}.\dfrac{1}{2017}\)
x = \(\dfrac{-1}{2019}\)
\(A=\left|2x-\dfrac{1}{3}\right|+1007\)
\(\left|2x-\dfrac{1}{3}\right|\ge0\)
\(\Rightarrow\left|2x-\dfrac{1}{3}\right|+1007\ge1007\)
Dấu "=" xảy ra khi:
\(\left|2x-\dfrac{1}{3}\right|=0\Rightarrow2x=\dfrac{1}{3}\Rightarrow x=\dfrac{1}{6}\)
\(\Rightarrow MIN_A=1007\) khi \(x=\dfrac{1}{6}\)
B tương tự
\(C=\left|2018-x\right|+\left|2017-x\right|\)
\(C=\left|2018-x\right|+\left|x-2017\right|\)
Áp dụng BĐT:
\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)
\(\Rightarrow C\ge\left|2018-x+x-2017\right|\)
\(C\ge1\)
Dấu "=" xảy ra khi:
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}2018-x\ge0\Rightarrow x\le2018\\x-2017\ge0\Rightarrow x\ge2017\end{matrix}\right.\\\left\{{}\begin{matrix}2018-x< 0\Rightarrow x>2018\\x-2017< 0\Rightarrow x< 2017\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow2017\le x\le2018\)
D tương tự
Ta có: \(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}=\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\)
\(\Rightarrow\left(\dfrac{x+1}{2017}+\dfrac{x+1}{2018}\right)-\left(\dfrac{x+1}{2019}+\dfrac{x+1}{2020}\right)=0\)
\(\Rightarrow\dfrac{x+1}{2017}+\dfrac{x+1}{2018}-\dfrac{x+1}{2019}-\dfrac{x+1}{2020}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}\right)=0\)
Vì \(\dfrac{1}{2017}>\dfrac{1}{2018}>\dfrac{1}{2019}>\dfrac{1}{2020}>0\) nên
\(\dfrac{1}{2017}+\dfrac{1}{2018}-\dfrac{1}{2019}-\dfrac{1}{2020}>0\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
x=-1