Question

    ……..

Answer 1

a: Khi \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\) thì \(A=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}-1}{\sqrt{\left(\sqrt{3}+1\right)^2}+2}\)

\(=\dfrac{\sqrt{3}+1-1}{\sqrt{3}+1+2}=\dfrac{\sqrt{3}}{3+\sqrt{3}}=\dfrac{1}{\sqrt{3}+1}=\dfrac{\sqrt{3}-1}{2}\)

b: A=1/2

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

=>\(2\sqrt{x}-2=\sqrt{x}+2\)

=>\(\sqrt{x}=4\)

=>x=16

c: A>1/3

=>A-1/3>0

=>\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}-\dfrac{1}{3}>0\)

=>\(\dfrac{3\sqrt{x}-3-\sqrt{x}-2}{3\left(\sqrt{x}+2\right)}>0\)

=>\(2\sqrt{x}-5>0\)

=>\(\sqrt{x}>\dfrac{5}{2}\)

=>x>25/4

d: A<1/căn x+2

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

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

=>\(\sqrt{x}-2< 0\)

=>\(\sqrt{x}< 2\)

=>0<=x<4

e: \(A+1=\dfrac{\sqrt{x}-1+\sqrt{x}+2}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+2}>0\)

=>A>-1

h: Để A nguyên thì \(\sqrt{x}-1⋮\sqrt{x}+2\)

=>\(\sqrt{x}+2-3⋮\sqrt{x}+2\)

=>\(\sqrt{x}+2\inƯ\left(-3\right)\)

=>\(\sqrt{x}+2=3\)

=>x=1(nhận)

f: Khi x>1 thì \(\sqrt{x}-1>0\)

=>A>0

\(\sqrt{x}-1< \sqrt{x}+2\) khi x>1

=>\(0< A< 1\)

=>\(A< \sqrt{A}\)

g: Khi 0<=x<1 thì \(\sqrt{x}-1< 0\)

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

=>A<|A|