4)
Với \(x=\dfrac{9}{16}\) TMĐK thay vào B ta có:
\(B=\dfrac{\sqrt{\dfrac{9}{16}}-3}{2}=\dfrac{\dfrac{3}{4}-3}{2}=\dfrac{\dfrac{-9}{4}}{2}=-\dfrac{9}{8}\)
5)
*Rút gọn A:
\(A=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}+11}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(A=\dfrac{2\sqrt{x}+14}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
Ta có \(M=A\cdot B=\dfrac{2\sqrt{x}+14}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2}=\dfrac{2\sqrt{x}+14}{2\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+7}{\sqrt{x}+3}\)
6)
Ta có \(M=\dfrac{\sqrt{x}+7}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3+4}{\sqrt{x}+3}=1+\dfrac{4}{\sqrt{x}+3}\)
Vì \(x\ge0\Rightarrow\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow\dfrac{1}{\sqrt{x}+3}\le\dfrac{1}{3}\Rightarrow\dfrac{4}{\sqrt{x}+3}\le\dfrac{4}{3}\Rightarrow1+\dfrac{4}{\sqrt{x}+3}\le\dfrac{7}{3}\)Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+3=3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)
Vậy Max(M) = \(\dfrac{7}{3}\Leftrightarrow x=0\)
