a: \(A=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
Khi x=4 thì \(A=\dfrac{4+2+1}{2\left(2+1\right)}=\dfrac{7}{2\cdot3}=\dfrac{7}{6}\)
b: B=1/3
=>\(\dfrac{\sqrt{x}}{\sqrt{x}+x}=\dfrac{1}{3}\)
=>\(\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{3}\)
=>căn x+1=3
=>căn x=2
=>x=4
c: \(B-1=\dfrac{1}{\sqrt{x}+1}-1=\dfrac{1-\sqrt{x}-1}{\sqrt{x}+1}=-\dfrac{\sqrt{x}}{\sqrt{x}+1}< 0\)
=>B<1

