a) \(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{9-\sqrt{9}+1}{\sqrt{9}-1}=\dfrac{9-3+1}{3-1}=\dfrac{7}{2}\)
b) \(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+2\left(\sqrt{x}-2\right)-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
c) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}>0\Leftrightarrow\sqrt{x}-1>0\left(do.\sqrt{x}+3>0\right)\)
\(\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)
\(B=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+1}{\sqrt{x}-1}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\)
Do \(\sqrt{x}>1\Leftrightarrow\sqrt{x}-1>0\)
Áp dụng BĐT Cauchy cho 2 số k âm:
\(B=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{\left(\sqrt{x}-1\right).\dfrac{1}{\sqrt{x}-1}}+1=2+1=3\)
Dấu "=" xảy ra \(\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow x=4\)
