Giải thích các bước giải:

a.Khi $x=16\to \sqrt{x}=4$

$\to A=\dfrac{2+4}2=3$

b.Ta có:

$B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}$

$\to B=\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)}$

$\to B=\dfrac{(\sqrt{x}-1)^2(\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}+\dfrac{2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}$

$\to B=\dfrac{(\sqrt{x}+1)^2(\sqrt{x}-1)+2\sqrt{x}+1}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}$

$\to B=\dfrac{\sqrt{x}(x+\sqrt{x}+1)}{\sqrt{x}(\sqrt{x}+1)(\sqrt{x}-1)}$

$\to B=\dfrac{x+\sqrt{x}+1}{(\sqrt{x}+1)(\sqrt{x}-1)}$

$\to B=\dfrac{x+\sqrt{x}+1}{x-1}$

c.Để $\dfrac{A}B>\dfrac43$

$\to \dfrac{x+\sqrt{x}+1}{x-1}>\dfrac43$

Vì $\dfrac43>0\to \dfrac{x+\sqrt{x}+1}{x-1}>0$

$\to x-1>0$ vì $x+\sqrt{x}+1>0$

$\to x>1$

$\to 3(x+\sqrt{x}+1)>4(x-1)$

$\to x-3\sqrt{x}-7<0$

$\to (\sqrt{x}-\dfrac32)^2-\dfrac{37}4<0$

$\to (\sqrt{x}-\dfrac32)^2<\dfrac{37}4$

$\to -\dfrac{\sqrt{37}}2<\sqrt{x}-\dfrac32<\dfrac{\sqrt{37}}2$

$\to \dfrac{3-\sqrt{37}}2<\sqrt{x}<\dfrac{3+\sqrt{37}}2$

Mà $0<x$

$\to 0<x<(\dfrac{3+\sqrt{37}}2)^2$