ĐKXĐ : \(x\ne0\)
Câu a :
\(A=\sqrt{\dfrac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\)
\(=\sqrt{\dfrac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^2+4x+4-8x}\)
\(=\sqrt{\dfrac{x^4+6x^2+9}{x^2}}+\sqrt{x^2-4x+4}\)
\(=\sqrt{\dfrac{\left(x^2+3\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)
\(=\left|\dfrac{x^2+3}{x}\right|+\left|x-2\right|\)
\(=\left|\dfrac{x^2}{x}+\dfrac{3}{x}\right|+\left|x-2\right|\)
Câu b :
Để \(A\in Z\Leftrightarrow\left|\dfrac{x^2}{x}+\dfrac{3}{x}\right|+\left|x-2\right|\in Z\)
\(\Leftrightarrow\dfrac{3}{x}\in Z\) ( Vì \(x^2⋮x\) )
\(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=-1\\x=1\\x=3\end{matrix}\right.\)
Vậy \(x=-3;x=-1;x=1;x=3\) thì A đạt giá trị nguyên .
Chúc bạn học tốt !!
