Question

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

tìm các giá trị nguyen của x để A có giá trị nguyên

Answer 1

\(ĐKXĐ:x\ne0\)

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

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

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

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

\(=\sqrt{\frac{\left(x^2+9\right)^2}{x^2}}+\sqrt{\left(x-2\right)^2}\)

\(=\frac{x^2+9}{\left|x\right|}+\left|x-2\right|=\frac{x^2}{\left|x\right|}+\frac{9}{\left|x\right|}+\left|x-2\right|\)

Vì \(x\inℤ\)\(\Rightarrow\frac{x^2}{\left|x\right|}\inℕ\), \(\left|x-2\right|\inℕ^∗\)

\(\Rightarrow\)Để A có giá trị nguyên thì \(\frac{9}{\left|x\right|}\inℕ^∗\)

\(\Rightarrow\left|x\right|\in\left\{1;3;9\right\}\)\(\Rightarrow x\in\left\{\pm1;\pm3;\pm9\right\}\)

Vậy \(x\in\left\{\pm1;\pm3;\pm9\right\}\)