Question

B=\(\dfrac{X}{\sqrt{X}-1}-\dfrac{2X-\sqrt{X}}{X-\sqrt{X}}\)

Tính B khi x=\(3+\sqrt{8}\)

Tính x để B >0

Answer 1

ĐK : \(x>0\) và \(x\ne1\)

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

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

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

Thay \(x=3+\sqrt{8}\) vào B ta được :

\(B=\sqrt{3+\sqrt{8}}-1=\sqrt{3+2\sqrt{2}}-1=\sqrt{2+2\sqrt{2}+1}-1=\sqrt{\left(\sqrt{2}+1\right)^2}-1=\sqrt{2}+1-1=\sqrt{2}\)

Để \(B>0\Leftrightarrow\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)