1: \(A=\left(\dfrac{2\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}-3}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{3\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
2: A=5/6
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{5}{6}\)
=>\(6\left(\sqrt{x}+1\right)=5\left(\sqrt{x}+3\right)\)
=>\(6\sqrt{x}+6=5\sqrt{x}+15\)
=>\(\sqrt{x}=9\)
=>x=81(nhận)
3: \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=1-\dfrac{2}{\sqrt{x}+3}\)
\(\sqrt{x}+3>=3\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{2}{\sqrt{x}+3}< =\dfrac{2}{3}\forall x\) thỏa mãn ĐKXĐ
=>\(-\dfrac{2}{\sqrt{x}+3}>=-\dfrac{2}{3}\forall x\) thỏa mãn ĐKXĐ
=>\(A=-\dfrac{2}{\sqrt{x}+3}+1>=-\dfrac{2}{3}+1=\dfrac{1}{3}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
\(A=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{3}{\sqrt{x}-3}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{3}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
b.
\(A=\dfrac{5}{6}\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{5}{6}\)
\(\Rightarrow6\left(\sqrt{x}+1\right)=5\left(\sqrt{x}+3\right)\)
\(\Rightarrow\sqrt{x}=9\Rightarrow x=81\)
c.
\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{3\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{1}{3}\)
Do \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\ge0;\forall x\ge0\Rightarrow A\ge\dfrac{1}{3}\)
\(A_{min}=\dfrac{1}{3}\) khi \(x=0\)
