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Côsi:
\(x+1=\left(x-2006\right)+2007\ge2\sqrt{2007}.\sqrt{x-2006}\)
\(x-1=\left(x-2007\right)+2006\ge2\sqrt{2006}.\sqrt{x-2007}\)
\(A\le\frac{1}{2\sqrt{2007}}+\frac{1}{2\sqrt{2006}}\)
Dấu bằng: \(\hept{\begin{cases}x-2006=2007\\x-2007=2006\end{cases}\Leftrightarrow x=2006+2007=4013}\)
a) ĐK : \(x\ge0\)
A = \(\frac{1}{\sqrt{x}+1}-\frac{3}{x\sqrt{x}+1}+\frac{1}{x-\sqrt{x}+1}\)
\(=\frac{x-\sqrt{x}+1-3+2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\cdot\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x-\sqrt{x}+1}\)
b) \(A=\frac{\sqrt{x}}{x-\sqrt{x}+1}=\frac{x-\sqrt{x}+1-x+2\sqrt{x}-1}{x-\sqrt{x}+1}=1-\frac{\left(\sqrt{x}-1\right)^2}{x-\sqrt{x}+1}\le1\)
=> Max A = 1
Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\)<=> x = 1
Vậy Max A = 1 <=> x = 1
a) ĐKXĐ: \(x\ge0;x\ne1\)
\(A=\left(1+\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right).\left(1-\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
b) \(A=1-x\le1\) ( vì \(x\ge0\) )
Vậy max A = 1 khi x = 0
a/ Ta có
P = \(\frac{1+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) - \(\frac{2+x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) - \(\frac{1+\sqrt{x}}{x+\sqrt{x}+1}\)
= \(\frac{-\sqrt{x}}{1+\sqrt{x}+x}\)
\(x\ge2017\)
\(A=\frac{\sqrt{x-2016}}{x-2016+2017}+\frac{\sqrt{x-2017}}{x-2017+2016}=\frac{1}{\sqrt{x-2016}+\frac{2017}{\sqrt{x-2016}}}+\frac{1}{\sqrt{x-2017}+\frac{2016}{\sqrt{x-2017}}}\)
\(A\le\frac{1}{2\sqrt{2017}}+\frac{1}{2\sqrt{2016}}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x-2016=2017\\x-2017=2016\end{matrix}\right.\) \(\Rightarrow x=4033\)