đk : x >= 0 ; x khác 1
\(C=\left(\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{x\sqrt{x}-1}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x-2\sqrt{x}+1}{x\sqrt{x}-1}:\dfrac{\sqrt{x}-1}{2}=\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có : \(x+\sqrt{x}+1=x+\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)
\(\Rightarrow C=\dfrac{2}{\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}>0\)
Vậy ta có đpcm
a:
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(C=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{x+\sqrt{x}+1}\)
b: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
nên C>0 với mọi x thỏa mãn ĐKXĐ