1: Thay x=16 vào B, ta được:
\(B=\dfrac{\sqrt{16}}{\sqrt{16}+2}=\dfrac{4}{4+2}=\dfrac{4}{6}=\dfrac{2}{3}\)
2: M=A:B
\(=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{2}{x-4}\right):\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2+2}{\left(\sqrt{x}-2\right)\cdot\sqrt{x}}=\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
3: M<1
=>M-1<0
=>\(\dfrac{\sqrt{x}+1-\left(\sqrt{x}-2\right)}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
Kết hợp ĐKXĐ, ta được: 0<x<4
