Question

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

a, rút gọn P

b, tìm x để P=\(\dfrac{9}{2}\)

Answer 1

Lời giải:

ĐK: \(x>0; x\neq 1\)

a)

\(P=\frac{(\sqrt{x})^3-1}{\sqrt{x}(\sqrt{x}-1)}-\frac{(\sqrt{x})^3+1}{\sqrt{x}(\sqrt{x}+1)}-\frac{x+1}{\sqrt{x}}\)

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

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

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

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

b) Ta thấy: \(\frac{(\sqrt{x}-1)^2}{\sqrt{x}}> 0,\forall x\neq 1>0\Rightarrow P=-\frac{(\sqrt{x}-1)^2}{\sqrt{x}}< 0\)

Do đó không tồn tại $x$ để \(P=\frac{9}{2}\)