Question

Cho biểu thức A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\). So sánh A và \(A^2\)

Answer 1

\(A^2-A=A\left(A-1\right)=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}=\dfrac{3\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-2\right)^2}>0\)

=>A^2>A