Question

\(Cho\) \(M=\dfrac{x-\sqrt{x}+3}{\sqrt{x}-3};N=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{5\sqrt{x}-9}{x-3\sqrt{x}}\) \(\left(x>0;x\ne9\right)\)

\(Q=M.N\) \(so\) \(sánh\) \(Q\) \(với\) \(2\)

Answer 1

*Rút gọn N

\(N=\dfrac{\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{5\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(N=\dfrac{x-\sqrt{x}-5\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(N=\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-3}{\sqrt{x}}\)

\(Q=M\cdot N=\dfrac{x-\sqrt{x}+3}{\sqrt{x}-3}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x-\sqrt{x}+3}{\sqrt{x}}\)

Xét hiệu Q-2 ta có: \(Q-2=\dfrac{x-\sqrt{x}+3}{\sqrt{x}}-2=\dfrac{x-3\sqrt{x}+3}{\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-\dfrac{3}{2}\right)^2+\dfrac{3}{4}}{\sqrt{x}}>0\)

\(\Rightarrow Q-2>0\Rightarrow Q>2\)