1) Thay \(x=9\) vào A ta có:
\(A=\dfrac{\sqrt{9}}{\sqrt{9}-1}=\dfrac{3}{3-1}=\dfrac{3}{2}\)
2) \(B=\dfrac{3x}{x-2\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{3x}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{3x-x+\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x+\sqrt{x}}{\left(\sqrt{x}-1\right)^2}\)
\(M=B:A\)
\(=\dfrac{2x+\sqrt{x}}{\left(\sqrt{x}-1\right)^2}:\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(M>1\) khi:
\(\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}>1\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+1}{\sqrt{x}-1}-1>0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+1-\sqrt{x}+1}{\sqrt{x}-1}>0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}-1}>0\)
Vì \(\sqrt{x}+2\ge2>0\forall x\left(tmđk\right)\) nên:
\(\sqrt{x}-1>0\)
\(\Leftrightarrow\sqrt{x}>1\)
\(\Leftrightarrow x>1\)
Vậy: ...
