Câu 2:
a: Thay x=25 vào A, ta được:
\(A=\dfrac{2\cdot\sqrt{25}+1}{\sqrt{25}+3}=\dfrac{2\cdot5+1}{5+3}=\dfrac{11}{8}\)
b: P=A:B
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}:\left(\dfrac{2x}{x-9}-\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}:\left(\dfrac{2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}:\dfrac{2x-\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{2x-x+3\sqrt{x}+x+4\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+1}{2x+7\sqrt{x}+3}\cdot\left(\sqrt{x}-3\right)\)
\(=\dfrac{\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(2\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-3\right)=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\)
c: \(P=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-6}{\sqrt{x}+3}=1-\dfrac{6}{\sqrt{x}+3}\)
\(\sqrt{x}+3>=3\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{6}{\sqrt{x}+3}< =\dfrac{6}{3}=2\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{6}{\sqrt{x}+3}>=-2\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{6}{\sqrt{x}+3}+1>=-1\forall x\) thỏa mãn ĐKXĐ
=>P>=-1 với mọi x thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
