Question

Cho biểu thức \(C=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2}{x-\sqrt{x}}\right):\dfrac{1}{1-\sqrt{x}}\left(x>0,x\ne1\right)\). Tìm Max C.

Answer 1

ĐKXĐ: x>0; x<>1

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

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

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

\(=-\left(\sqrt{x}+\dfrac{2}{\sqrt{x}}\right)< =-\left(2\cdot\sqrt{\sqrt{x}\cdot\dfrac{2}{\sqrt{x}}}\right)=-2\sqrt{2}\) với mọi x thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=2\)

=>x=2