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\(\frac{\sqrt{\left(x-2017\right)2019}}{\sqrt{2019}\left(x+2\right)}+\frac{\sqrt{\left(x-2018\right)2018}}{\sqrt{2018}x}\le\frac{x-2017+2019}{2\sqrt{2019}\left(x+2\right)}+\frac{x-2018+2018}{2\sqrt{2018}x}\)
\(=\frac{1}{2\sqrt{2019}}+\frac{1}{2\sqrt{2018}}\)
''='' khi x=4036
\(\frac{x-2017}{2018}-\frac{x-2018}{2017}=\frac{2017}{x-2018}-\frac{2018}{x-2017}\)
\(\Leftrightarrow\)\(\frac{2017\left(x-2017\right)-2018\left(x-2018\right)}{2017.2018}=\frac{2017\left(x-2017\right)-2018\left(x-2018\right)}{\left(x-2017\right)\left(x-2018\right)}\)
Do \(2017\left(x-2017\right)-2018\left(x-2018\right)\ne0\) nên \(\left(x-2017\right)\left(x-2018\right)=2017.2018\)
\(\Leftrightarrow\)\(x^2-4035x+2017.2018=2017.2018\)
\(\Leftrightarrow\)\(x\left(x-4035\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\left(l\right)\\x=4035\left(n\right)\end{cases}}\)
Vậy x = 4035
Ta có: \(A=\left(x+y\right).1=\left(x+y\right).\left(\frac{2017}{x}+\frac{2018}{y}\right)=2017+2018.\frac{x}{y}+2017.\frac{y}{x}+2018\)
\(\Leftrightarrow A=4035+2017\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{x}{y}\ge4035+2017.2+\frac{x}{y}\)
\(\Leftrightarrow A\ge8069+\frac{x}{y}\)
Dấu " = " xảy ra \(\Leftrightarrow\frac{x}{y}=\frac{y}{x}\Leftrightarrow x^2=y^2\Leftrightarrow x=y=4035\)( thỏa đề bài )
\(\Leftrightarrow minA=8069+1=8070\)
\(Q=\frac{2017}{x-8\sqrt{x}+2018}=\frac{2017}{\left(\sqrt{x}-4\right)^2+2002}\)
ta có \(\left(\sqrt{x}-4\right)^2\ge0\)
\(Q\le\frac{2017}{2002}\)
dấu "=" xảy ra khi \(x=16\)
\(MAX:Q=\frac{2017}{2002}\)