Question

nhanh hộ mình nhé mình xin cảm ơn

Answer 1

a: \(A=\sqrt{3}+1-\sqrt{3}+1=2\)

\(B=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{x+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)

b: Để A>2B thì A-2B>0

=>\(\dfrac{2\sqrt{x}-4-\sqrt{x}+1}{\sqrt{x}-2}>0\)

\(\Leftrightarrow\dfrac{\sqrt{x}-3}{\sqrt{x}-2}>0\)

=>x>9 hoặc 0<=x<4