Question

A= \(\sqrt{\frac{\left(x^2-3^2\right)+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x^2}\)

Rút gọn A

Tìm những giá trị nguyên để A nguyên

Answer 1

Lời giải:
ĐK: $x\neq 0$

Ta có:
\(A=\sqrt{\frac{(x^2-3)^2+12x^2}{x^2}}+\sqrt{(x^2+2)^2-8x^2}\)

\(=\sqrt{\frac{x^4-6x^2+9+12x^2}{x^2}}+\sqrt{x^4+4x^2+4-8x^2}\)

\(=\sqrt{\frac{x^4+6x^2+9}{x^2}}+\sqrt{x^4-4x^2+4}\)

\(=\sqrt{\frac{(x^2+3)^2}{x^2}}+\sqrt{(x^2-2)^2}\)

\(=|\frac{x^2+3}{x}|+|x^2-2|\)

Để $A$ nguyên thì \(|\frac{x^2+3}{x}|+|x^2-2|\in\mathbb{Z}\)

\(\Leftrightarrow |\frac{x^2+3}{x}|\in\mathbb{Z}\)

\(\Leftrightarrow x^2+3\vdots x\)

\(\Leftrightarrow 3\vdots x\Rightarrow x\in \left\{\pm 1;\pm 3\right\}\)

Vậy.........