Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

\(A=\frac{\left|x-2016\right|+2017}{\left|x-2016\right|+2018}\)

\(A=\frac{\left|x-2016\right|+2018}{\left|x-2016\right|+2018}-\frac{1}{\left|x-2016\right|+2018}\)

\(A=1-\frac{1}{\left|x-2016\right|+2018}\ge1-\frac{1}{2018}=\frac{2017}{2018}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\left|x-2016\right|=0\)

\(\Leftrightarrow\)\(x=2016\)

Vậy GTNN của \(A\) là \(\frac{2017}{2018}\) khi \(x=2016\)

Chúc bạn học tốt ~ 

300m2

ta có x^2+y^2-6x+18+6y=0

(x-3)^2+(y+3)^2=0

x=3 và y=-3 thay vào biểu thức A bạn sẽ tính dc kq

Nguyễn Tiến Đạt

a)\(|3x-5|=|x+2|\)

=> Ta có 2 trường hợp

*) TH1: 3x-5=x+2

=>3x-x=2+5

=>2x=7

=>x=7:2\(\Rightarrow x=\frac{7}{2}\)

*)TH2: -3x+5=x+2

\(\Rightarrow5-3x=x+2\)

\(\Rightarrow5-2=x+3x\)

\(\Rightarrow3=4x\)

\(\Rightarrow x=3:4\Rightarrow x=\frac{3}{4}\)

Vậy \(x\in\left\{\frac{7}{2};\frac{3}{4}\right\}\)

\(\frac{x-3}{2017}-\frac{x-2}{2018}=\frac{x-2018}{2}+\frac{x-2017}{3}\)

\(\Leftrightarrow\frac{x-3}{2017}-1-\frac{x-2}{2018}-1=\frac{x-2018}{2}-1+\frac{x-2017}{3}-1\)

\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}=\frac{x-2020}{2}+\frac{x-2020}{3}\)

\(\Leftrightarrow\frac{x-2020}{2017}-\frac{x-2020}{2018}-\frac{x-2020}{2}-\frac{x-2020}{3}=0\)

\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2}-\frac{1}{3}\right)=0\)

\(\Leftrightarrow x-2020=0\Leftrightarrow x=2020\)