Question

Answer 1

a: ĐKXĐ: x>0; x<>1

\(Q=\dfrac{x+\sqrt{x}+\sqrt{x}}{x-1}:\dfrac{2\sqrt{x}+2-2+x}{x\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{x}{\sqrt{x}-1}\)

b: Q>2

=>Q-2>0

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

=>căn x-1>0

=>x>1

c: \(Q=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\)

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

Dấu = xảy ra khi căn x-1=1

=>căn x=2

=>x=4