Question

Cho biểu thức P(x) = \(\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{x}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\), với x>0 và x ≠ 1. Rút gọn P(x) và tìm x để Q(x) = \(\frac{2\sqrt{x}}{P\left(x\right)}\)nhận giá trị nguyên

Answer 1

Ý 1:

\(P\left(x\right)=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{x}+\frac{2\left(x-1\right)}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x+1}\right)\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)\)

\(=x-\sqrt{x}-2\sqrt{x}-1+2\sqrt{x}+2\)

\(=x-\sqrt{x}+1\)

Ý 2:

\(P\left(x\right)=\frac{2\sqrt{x}}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}-1}\)

Vì: \(\sqrt{x}+\frac{1}{\sqrt{x}}-1>1\) nên:

\(\Rightarrow\sqrt{x}+\frac{1}{\sqrt{x}}-1=2\)

Để \(Q\) nguyên \(\Leftrightarrow\sqrt{x}+\frac{1}{\sqrt{x}}-1\inƯ\left(2\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1=\frac{7+3\sqrt{5}}{2}\\x_2=\frac{7-3\sqrt{5}}{2}\end{matrix}\right.\)

Vậy .................................