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Ta có:

\(\sqrt{x^2-2018x+2018}+\sqrt{x^2-1009x+1009}=2x\)

\(\Leftrightarrow x-\sqrt{\left(2018x-2018\right)}+x-\sqrt{\left(1009x-1009\right)}=2x\)

\(\Leftrightarrow2x-\sqrt{\left(2018x-2018\right)}-\sqrt{\left(1009x-1009\right)}=2x\)

\(\Leftrightarrow\sqrt{\left(2018x\right)-2018}+\sqrt{\left(1009x-1009\right)}=0\)

\(\Leftrightarrow\sqrt{\left(2018x-2018\right)}=\sqrt{\left(1009x-1009\right)}=0\)

\(\Leftrightarrow2018x-2018=1009x-1009=0\Leftrightarrow x=1\)

<=> \(x^4\left(\sqrt{x+3}-2\right)\)\(+2018\left(x-1\right)=0\)

<=>\(x^4\left(\dfrac{x+3-4}{\sqrt{x+3}+2}\right)+2018\left(x-1\right)=0\)

<=>\(x^{\text{4}}\left(\dfrac{x-1}{\sqrt{x+3}+2}\right)+2018\left(x-1\right)=0\)

<=>\(\left(x-1\right)\left(\dfrac{x^4}{\sqrt{x+3}+2}+2018\right)=0\)

=>x-1=0 <=>x=1

Ta có: \(\left|x\right|+\left|x+1\right|+\cdots+\left|x+2018\right|=x^2+2018x-2019\)

=>\(x^2+2018x-2019\ge0\)

=>(x+2019)(x-1)>=0

=>\(\left[\begin{array}{l}x\ge1\\ x\le-2019\end{array}\right.\)

TH1: x>=1

=>|x|=x; |x+1|=x+1; ...; |x+2018|=x+2018

Ta có: \(\left|x\right|+\left|x+1\right|+\cdots+\left|x+2018\right|=x^2+2018x-2019\)

=>\(x^2+2018x-2019=x+x+1+\cdots+x+2018\)

=>\(x^2+2018x-2019=2019x+\left(1+2+\cdots+2018\right)\)

=>\(x^2-x-2019-2018\cdot\frac{2019}{2}=0\)

=>\(x^2-x-2019-2019\cdot1009=0\)

=>\(x^2-x-2019\cdot1010=0\)

\(\Delta=\left(-1\right)^2-4\cdot1\cdot\left(-2019\cdot1010\right)=8156761\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{1-\sqrt{8156761}}{2\cdot1}=\frac{1-\sqrt{8156761}}{2}\left(loại\right)\\ x=\frac{1+\sqrt{8156761}}{2}\left(nhận\right)\end{array}\right.\) (2)

TH2: x<=-2019

=>x+1<=-2018<0; x+2<=-2017<0; ...; x+2018<=-2019+2018=-1<0

=>|x|=-x; |x+1|=-(x+1); |x+2|=-(x+2);...;|x+2018|=-(x+2018)

Ta có: \(\left|x\right|+\left|x+1\right|+\cdots+\left|x+2018\right|=x^2+2018x-2019\)

=>\(x^2+2018x-2019=-\left(x+x+1+\cdots+x+2018\right)\)

=>\(x^2+2018x-2019=-\left\lbrack2019x+\left(1+2+\cdots+2018\right)\right\rbrack\)

=>\(x^2+2018x-2019=-2019x-2018\cdot\frac{2019}{2}=-2019x-2037171\)

=>\(x^2+4037x+2035152=0\)

\(\Delta=4037^2-4\cdot1\cdot2035152=8156761>0\)

Do đó: Phương trình có hai nghiệm phân biệt là:

\(\left[\begin{array}{l}x=\frac{-4037-\sqrt{8156761}}{2}\simeq-3446,502\left(nhận\right)\\ x=\frac{-4037+\sqrt{8156761}}{2}\simeq-590,498\left(loại\right)\end{array}\right.\) (1)

Từ (1),(2) suy ra phương trình có hai nghiệm

ĐK: \(x\ge\frac{2017}{2018}\)

\(pt\Leftrightarrow2017\sqrt{2017x-2016}-2017+\sqrt{2018x-2017}-1=0\)

\(\Leftrightarrow2017\frac{2017\left(x-1\right)}{\sqrt{2017x-2016}+1}+\frac{2018\left(x-1\right)}{\sqrt{2018x-2017}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}\right)=0\)

Dễ thấy với \(x\ge\frac{2017}{2018}\Rightarrow\)\(\frac{2017^2}{\sqrt{2017x-2016}+1}+\frac{2018}{\sqrt{2018x-2017}+1}>0\)

\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)

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

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