Chiều mk lm cho

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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

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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