Question

Answer 1

1: Thay x=9 vào A, ta được:

\(A=\dfrac{3\cdot3-4}{3+1}=\dfrac{5}{4}\)

2: \(B=\dfrac{2x+\sqrt{x}-4-x+4+\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

\(P=A\cdot B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\cdot\dfrac{3\sqrt{x}-4}{\sqrt{x}+1}=\dfrac{3\sqrt{x}-4}{\sqrt{x}-2}\)

3: Để P>=2 thì P-2>=0

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

\(\Leftrightarrow\sqrt{x}-2>0\)

hay x>4