1: Thay x=9 vào A, ta được:
\(A=\dfrac{3+4}{3-1}=\dfrac{7}{2}\)
2: \(B=\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{2}{\sqrt{x}+3}\)
\(=\dfrac{3\sqrt{x}+1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{1}{\sqrt{x}-1}\)
3: \(\dfrac{A}{B}>=\dfrac{x}{4}+5\)
=>\(\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\dfrac{1}{\sqrt{x}-1}>=\dfrac{x}{4}+5=\dfrac{x+20}{4}\)
=>\(\sqrt{x}+4>=\dfrac{x+20}{4}\)
=>\(x+20< =4\sqrt{x}+16\)
=>\(x-4\sqrt{x}+4< =0\)
=>\(\left(\sqrt{x}-2\right)^2< =0\)
=>\(\sqrt{x}-2=0\)
=>x=4(nhận)
