Question

so sánh P với 1 biết P=căn x- 2 trên 2 căn x +1

Answer 1

\(P=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}\) với \(x\ge0\)

Ta có: \(P-1=\dfrac{\sqrt{x}-2}{2\sqrt{x}+1}-1=\dfrac{\sqrt{x}-2-2\sqrt{x}-1}{2\sqrt{x}+1}=-\dfrac{\sqrt{x}+3}{2\sqrt{x}+1}\)

Do \(\sqrt{x}\ge0;\forall x\ge0\)

\(\Rightarrow\dfrac{\sqrt{x}+3}{2\sqrt{x}+1}>0\Rightarrow-\dfrac{\sqrt{x}+3}{2\sqrt{x}+1}< 0\)

\(\Rightarrow P-1< 0\Rightarrow P< 1\)