Mình chịu

\(A=\sqrt{2\sqrt{5}+6}-\sqrt{\left(\sqrt{5}-1\right)^2}+2018=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{5}+2019=2020\)

Ma \(2020\in Z\)

Suy ra:\(A\in Z\)

\(-2A=2x^2+2y^2-2xy-4x-4y\)

\(=\left(x^2-2xy+y^2\right)+\left(x^2-4x+4\right)+\left(y^2-4y+4\right)-8\)

\(=\left(x-y\right)^2+\left(x-2\right)^2+\left(y-2\right)^2-8\ge-8\)

=> \(A\le4\)

"=" xảy ra <=> x=y=2

Vậy max A=4 tại x=y=2

ĐK: \(\hept{\begin{cases}x^2-x+2\ge0\\2x^2+4x\ge0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\ge0\\x\le-2\end{cases}}\)

Ta có: \(VP=2\sqrt{x^2-x+2}-\sqrt{2x^2+4x}=\frac{2\left(x-2\right)^2}{2\sqrt{x^2-x+2}+\sqrt{2x^2+4x}}\ge0\)

=> \(VP=x-2\ge0\Rightarrow x\ge2\)

phương trình tương đương:

 \(2x-2\sqrt{x^2-x+2}+\sqrt{2x^2+4x+2}-x-2=0\)

\(\Leftrightarrow\frac{2\left(x-2\right)}{2x+2\sqrt{x^2-x+2}}+\frac{x^2-4}{\sqrt{2x^2+4x+2}+x+2}=0\)

\(\text{​​}\Leftrightarrow\left(x-2\right)\left[\frac{2}{2x+2\sqrt{x^2-x+2}}+\frac{x+2}{\sqrt{2x^2+4x+2}+x+2}\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\frac{2}{2x+2\sqrt{x^2-x+2}}+\frac{x+2}{\sqrt{2x^2+4x+2}+x+2}\end{cases}=0}\left(1\right)\)

(1) vô nghiệm vì x >=2 

Vậy pt <=> x=2

\(A=\frac{1}{\sqrt{x^2}-2x-1}=\frac{1}{\left|x\right|-2x-1}\)

* Xét \(x\ge0\)thì \(\left|x\right|=x\)

Lúc đó\(A=\frac{1}{x-2x-1}=\frac{1}{-x-1}\)

A có nghĩa\(\Leftrightarrow-x-1\ne0\Leftrightarrow x\ne-1\)

* Xét \(x< 0\)thì \(\left|x\right|=-x\)

Lúc đó\(A=\frac{1}{-x-2x-1}=\frac{1}{-3x-1}\)

A có nghĩa\(\Leftrightarrow-3x-1\ne0\Leftrightarrow x\ne\frac{-1}{3}\)

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