1: Thay x=16 vào A, ta được:
\(A=\dfrac{16+7}{\sqrt{16}}=\dfrac{23}{4}\)
2: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}-\dfrac{2x-\sqrt{x}-3}{x-9}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)-2x+\sqrt{x}+3}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-\sqrt{x}-3-2x+\sqrt{x}+3}{x-9}\)
\(=\dfrac{x+3\sqrt{x}}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
3: \(P=A+\dfrac{1}{B}=\dfrac{x+7}{\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}=\dfrac{x+\sqrt{x}+4}{\sqrt{x}}=\sqrt{x}+1+\dfrac{4}{\sqrt{x}}\)
=>\(P>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{4}{\sqrt{x}}}+1=2\cdot2+1=5\)
Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=4\)
=>x=4(nhận)
