Question

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\) với x≥ 0;x khác 4 tìm x đẻ 1)A=2 2)A<1 3)A< 1/3 4)A=\(\sqrt{x}\)

  

Answer 1

1: A=2

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

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

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

=>x=25

2: A<1

=>A-1<0

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

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

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

=>0<=x<4

3: A<1/3

=>A-1/3<0

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

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

=>\(\dfrac{2\sqrt{x}+5}{3\left(\sqrt{x}-2\right)}< 0\)

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

=>0<=x<4

4:
A=căn x

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

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

=>\(\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{13}}{2}\left(nhận\right)\\\sqrt{x}=\dfrac{3-\sqrt{13}}{2}\left(loại\right)\end{matrix}\right.\)

=>\(x=\dfrac{11+3\sqrt{13}}{2}\)