a: Khi x=9 thì \(A=\dfrac{\sqrt{9}+2}{9-25}=\dfrac{3+2}{-16}=\dfrac{-5}{16}\)
b: \(B=\dfrac{3}{\sqrt{x}+5}-\dfrac{20-2\sqrt{x}}{25-x}\)
\(=\dfrac{3}{\sqrt{x}+5}-\dfrac{2\sqrt{x}-20}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\dfrac{3\sqrt{x}-15-2\sqrt{x}+20}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)
\(=\dfrac{\sqrt{x}+5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\dfrac{1}{\sqrt{x}-5}\)
c: P=A:B
\(=\dfrac{\sqrt{x}+2}{x-25}:\dfrac{1}{\sqrt{x}-5}\)
\(=\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\sqrt{x}-5}{1}=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}=\dfrac{\sqrt{x}+5-3}{\sqrt{x}+5}=1-\dfrac{3}{\sqrt{x}+5}\)
\(\sqrt{x}+5>=5\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{3}{\sqrt{x}+5}< =\dfrac{3}{5}\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{3}{\sqrt{x}+5}>=-\dfrac{3}{5}\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{3}{\sqrt{x}+5}+1>=\dfrac{2}{5}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
