Question

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

a) Rút gọn \(P\)

b) Tìm \(x\in Z\)để \(P\in Z\)

Answer 1

\(P=\sqrt{\frac{\left(x^2-3\right)^2+12x^2}{x^2}}+\sqrt{\left(x+2\right)^2-8x}\) Đk \(x\ne0\)

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

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

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

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

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

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

b, \(P=\frac{2x^2-2x+3}{x}=2x-2+\frac{3}{x}\)

Để \(P\in z\)thì \(x\inƯ\left(3\right)=\left(-3;-1;1;3\right)\)