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\(A^2=\left(2\sqrt{x-4}+\sqrt{8-x}\right)^2\le\left(2^2+1^2\right)\left(x-4+8-x\right)=20..\)
\(A\le2\sqrt{5}..\)
Ta có: \(R=\sqrt{x^2-2x+1}+\sqrt{x^2+2x+1}\)
\(=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+1\right)^2}\)
\(=\left|x-1\right|+\left|x+1\right|\)
Ta có: \(-1\le x\le1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\le1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1\ge0\\x-1\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|x+1\right|=x+1\\\left|x-1\right|=1-x\end{matrix}\right.\)
\(\Leftrightarrow R=x+1+1-x=2\)
\(A^2=\left(\sqrt{13}.\sqrt{13x^2-13x^4}+3\sqrt{3}.\sqrt{3x^2+3x^4}\right)^2\)
\(\Rightarrow A^2\le\left(13+27\right)\left(16x^2-10x^4\right)=40\left[\frac{32}{5}-10\left(x^2-\frac{4}{5}\right)^2\right]\le256\)
\(\Rightarrow A\le16\Rightarrow A_{max}=16\) khi \(x^2=\frac{4}{5}\)
P=\(\sqrt{\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1}\)
=\(\sqrt{\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1}\)
=\(\sqrt{x-\sqrt{x}-x-\sqrt{x}+x+1}\)
=\(\sqrt{x-2\sqrt{x}+1}\)
=\(\sqrt{\left(\sqrt{x}-1\right)^2}\)
=\(\sqrt{x}-1\)