Question

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

a, Rút gọn A b, Tìm x nguyên để A đạt giá trị nguyên

Answer 1

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

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

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

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

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

\(=\left[{}\begin{matrix}\frac{x^2+3}{x}+x-2\left(x\ge2\right)\\\frac{x^2+3}{x}+2-x\left(x< 2\right)\end{matrix}\right.\)

\(=\left[{}\begin{matrix}\frac{x^2+3+x^2-2x}{x}\\\frac{x^2+3+2x-x^2}{x}\end{matrix}\right.\)

\(=\left[{}\begin{matrix}\frac{2x^2-2x+3}{x}\\\frac{2x+3}{x}\end{matrix}\right.\)