a)
Thay x = 16 TMĐK vào A ta có:
\(A=\dfrac{16+3}{\sqrt{16}+3}=\dfrac{19}{4+3}=\dfrac{19}{7}\)
b) \(B=\left[\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(B=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(B=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
c) \(P=\dfrac{A}{B}=\dfrac{x+3}{\sqrt{x}+3}\div\dfrac{\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{x+3}{\sqrt{x}+1}\)
\(P=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1}{\sqrt{x}+1}+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)
Vì \(x\ge0\Rightarrow\sqrt{x}+1>0\Rightarrow\dfrac{4}{\sqrt{x}+1}>0\)
Áp dụng BĐT Cauchy ta có:
\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}\ge2\sqrt{\left(\sqrt{x}+1\right)\left(\dfrac{4}{\sqrt{x}+1}\right)}=2\sqrt{4}=4\)
\(\Rightarrow\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\ge2\)
Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+1=\dfrac{4}{\sqrt{x}+1}\Leftrightarrow\left(\sqrt{x}+1\right)^2=4\Leftrightarrow x=1\)
Vậy Min P = 2 \(\Leftrightarrow x=1\)