Question

Cho: P= \(\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

a) Rút gọn

b) So sánh P vs \(\frac{1}{3}\)

Answer 1

a) \(P=\left(\frac{x+3}{x-9}+\frac{1}{\sqrt{x}+3}\right):\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(P=\left(\frac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right)\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(P=\frac{x+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}}\)

\(P=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\cdot\sqrt{x}}\)

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

b) Xét \(\frac{1}{3}\)có \(1< 3\)và \(\sqrt{x}\ge0\)

\(\Rightarrow\frac{1}{3}< \frac{1+\sqrt{x}}{3+\sqrt{x}}=P\)