Question

So sánh P=\(\dfrac{x-\sqrt{x}+1}{\sqrt{x}}\) với 1

Answer 1

ĐKXĐ: \(x>0\)

\(P=\sqrt{x}-1+\dfrac{1}{\sqrt{x}}=\sqrt{x}+\dfrac{1}{\sqrt{x}}-1\ge2\sqrt{\sqrt{x}.\dfrac{1}{\sqrt{x}}}-1=2-1=1\)

\(\Rightarrow P\ge1\) \(\forall x>0\) . Dấu "=" xảy ra khi x=1

Answer 2

ĐKXĐ: x > 0\(P-1=\dfrac{x-\sqrt{x}+1}{\sqrt{x}}-1=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\)

=> P ≥ 1

Dấu "=" xảy ra <=> x = 1