a: Thay x=16 vào A, ta được:
\(A=\dfrac{16+3}{4+3}=\dfrac{19}{7}\)
b: \(B=\left(\dfrac{x+3\sqrt{x}-2}{x-9}-\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\left(\dfrac{x+3\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{1}{\sqrt{x}+3}\right)\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{x+3\sqrt{x}-2-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
c: \(P=\dfrac{A}{B}=\dfrac{x+3}{\sqrt{x}+3}:\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(=\dfrac{x+3}{\sqrt{x}+3}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}+1}\)
\(=\dfrac{x+3}{\sqrt{x}+1}=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}\)
\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2\)
\(\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}>=2\cdot\sqrt{\left(\sqrt{x}+1\right)\cdot\dfrac{4}{\sqrt{x}+1}}=4\)
=>\(P=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2>=4-2=2\)
Dấu '=' xảy ra khi \(\sqrt{x}+1=\sqrt{4}=2\)
=>x=1(nhận)
