Question

cho \(B\)= \(\frac{\sqrt{x}+1}{\sqrt{x-1}}\)

so sánh \(Bvới\sqrt{B}với;x>1\)

Answer 1

\(2\sqrt{x}+2>0 \Rightarrow x+2\sqrt{x}+1>x-1\)
\(\Rightarrow\left(\sqrt{x}+1\right)^2>x-1\Rightarrow\sqrt{x}+1>\sqrt{x-1}\)
\(\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x-1}}>1hayB>1\)
Vì \(B>1\Rightarrow B>\sqrt{B}\)

Answer 2

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Answer 3

Ta có

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

Vì x>1  =>   \(\sqrt{x}>0\)

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

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

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

=>\(B>\sqrt{B}\)