a, ĐKXĐ :\(\left\{{}\begin{matrix}x\ne0\\x^2+4x+4\ge8x\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne0\\x^2-4x+4\ge0\end{matrix}\right.\)
=> \(x\ne0\)
Ta có : \(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
=> \(P=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)
=> \(P=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)
=> \(P=\sqrt{\left(\frac{x^2+3}{x}\right)^2}+\sqrt{\left(x-2\right)^2}\)
=> \(P=\frac{x^2+3}{\left|x\right|}+\left|x-2\right|\)
TH1 : \(\left\{{}\begin{matrix}x\ge0\\x-2\ge0\end{matrix}\right.\)
=> \(x\ge2\)
=> \(\frac{x^2+x^2-2x+3}{x}=\frac{2x^2-2x+3}{x}=\frac{2x\left(x-1\right)+3}{x}=2\left(x-1\right)+\frac{3}{x}\)
=> \(x\inƯ_{\left(3\right)}\)
=> \(x=\left\{1,-1,3,-3\right\}\)
TH2 : \(\left\{{}\begin{matrix}x\le0\\x-2\le0\end{matrix}\right.\)
=> \(x\le0\)
=> \(\frac{x^2+3-x\left(x-2\right)}{-x}=\frac{x^2-x^2+2x+3}{-x}=\frac{-2x-3}{x}=-2-\frac{3}{x}\)
=> \(x\inƯ_{\left(3\right)}\)
=> \(x=\left\{1,-1,3,-3\right\}\)
TH3 : \(\left\{{}\begin{matrix}x\ge0\\x-2\le0\end{matrix}\right.\)
=> \(0\le x\le2\)
=> \(\frac{x^2+3}{x}+2-x=\frac{x^2+3+2x-x^2}{x}=2+\frac{3}{x}\)
=> \(x\inƯ_{\left(3\right)}\)
=> \(x=\left\{1,-1,3,-3\right\}\)
TH4 : \(\left\{{}\begin{matrix}x\le0\\x-2\ge0\end{matrix}\right.\)
=> Vô lý .
Vậy để giá trị của x để P nguyên thì \(x=\left\{1,-1,3,-3\right\}\) .
a) $ĐKXĐ : x \neq 0$
Ta có :
\(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
\(=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)
\(=\sqrt{\frac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)
\(=\left|\frac{x^2+3}{x}\right|+\left|x-2\right|\)
Với \(x< 0\) thì : \(P=\frac{x^2+3}{-x}+2-x=\frac{2x^2+3-2x}{-x}\)
Với \(0< x< 2\) thì : \(P=\frac{x^2+3}{x}+2-x=\frac{2x+3}{x}\)
Với \(x\ge2\) thì : \(P=\frac{x^2+3}{x}+x-2=\frac{2x^2+3-2x}{x}\)
Để $P$ nguyên thì : $x^2+3 \vdots x$
$\to x \in Ư(3)$
