Question

CM  \(\frac{\sqrt{x}-x+1}{x-1}< 0\)

Answer 1

đk:\(x\ge0:x\ne1\)

Ta có; \(\frac{\sqrt{x}-x+1}{x-1}=\frac{-\sqrt{x}\left(\sqrt{x}-1\right)+1}{\left(\sqrt{x-1}\right)\left(\sqrt{x}+1\right)}=-\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{x-1}\le\frac{1}{x-1}\left(do\sqrt{x}>0\right)\)

\(\Rightarrow\frac{\sqrt{x}-x+1}{x-1}< 0\left(dpcm\right)\)