giúp mk với

giúp mk với

\(x\ge2017\)

\(A=\frac{\sqrt{x-2016}}{x-2016+2017}+\frac{\sqrt{x-2017}}{x-2017+2016}=\frac{1}{\sqrt{x-2016}+\frac{2017}{\sqrt{x-2016}}}+\frac{1}{\sqrt{x-2017}+\frac{2016}{\sqrt{x-2017}}}\)

\(A\le\frac{1}{2\sqrt{2017}}+\frac{1}{2\sqrt{2016}}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-2016=2017\\x-2017=2016\end{matrix}\right.\) \(\Rightarrow x=4033\)

\(A=\left[\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\frac{\sqrt{x}-2}{\sqrt{x}+2}\right].\left[\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}+\sqrt{x}+4\right]\) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

\(=\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}+2}.\left(x+5\right)\)

\(=\frac{x+5}{\sqrt{x}+2}\)

\(=\frac{2\left(\sqrt{x}+2\right)}{\sqrt{x}+2}+\frac{x-2\sqrt{x}+1}{\sqrt{x}+2}\)

\(=2+\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+2}\ge2\)

Dấu '=' xảy ra khi \(x=1\)

Vậy \(A_{min}=2\) khi \(x=1\)

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\)

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

\(=\left(x-\frac{2\sqrt{x}}{2}+\frac{1}{4}\right)-\frac{1}{4}=\left(\sqrt{x}-\frac{1}{4}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

Vậy GTNN là \(\frac{-1}{4}\)đạt được khi x = \(\frac{1}{4}\)

Ai trả lời nhanh và chính xác mình k