Nhận xét : ( x + y - 3 )^2018 >=0 và 2018.(2x-4)^2020 >= 0

=> (x+y-3)^2018 + 2018.(2x-4)^2020 >=0 

Dấu = xảy ra khi : x + y - 3 = 0 và 2x - 4 = 0 => x = 2 và y = 1

Thay vào bt S :

S = ( 2 - 1)^2019 + (2-1)^2019

= 1^2019 + 1^2019 = 2

em cảm ơn

Ta có: \(\left|x-1\right|+\left|x-2020\right|=\left|x-1\right|+\left|2020-x\right|\ge\left|x-1+2020-x\right|=2019\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(2020-x\right)\ge0\)\(\Leftrightarrow1\le x\le2020\)

Vì \(\hept{\begin{cases}\left|x-30\right|\ge0\\\left|y-4\right|\ge0\\\left|z-1975\right|\ge0\end{cases}}\forall x,y,z\)\(\Rightarrow\left|x-1\right|+\left|x-30\right|+\left|y-4\right|+\left|z-1975\right|+\left|x-2020\right|\ge2019\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}x-30=0\\y-4=0\\z-1975=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=30\\y=4\\z=1975\end{cases}}\)

So sánh \(x=30\)với điều kiện \(1\le x\le2020\)ta được x thoả mãn

Vậy \(x=30\); \(y=4\); \(z=1975\)

a) \(P=\left|x-2016\right|+\left|x-2017\right|+\left|x-2018\right|\)

*TH1: \(x< 2016\):

\(P=2016-x+2017-x+2018-x=6051-3x>6051-3\cdot2016=3\)

*TH2: \(2016\le x< 2017\):

\(P=x-2016+2017-x+2018-x=2019-x>2019-2017=2\)

*TH3: \(2017\le x< 2018\):

\(P=x-2016+x-2017+2018-x=x-2015\ge2017-2015=2\)(Dấu "=" xảy ra khi x = 2017)

*TH4: \(x\ge2018\):

\(P=x-2016+x-2017+x-2018=3x-6051\ge3\cdot2018-6051=3\)(Dấu "=" xảy ra khi x = 2018)

Vậy GTNN của P là 2 khi x = 2017.

b) \(x-2xy+y-3=0\)

\(\Leftrightarrow x\left(1-2y\right)+y-\frac{1}{2}-\frac{5}{2}=0\)

\(\Leftrightarrow2x\left(\frac{1}{2}-y\right)-\left(\frac{1}{2}-y\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(\frac{1}{2}-y\right)=\frac{5}{2}\)

\(\Leftrightarrow\left(2x-1\right)\left(1-2y\right)=5\)