Question

Answer 1

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

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

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

b Để D=0,5 thì \(x-\sqrt{x}+1=\dfrac{1}{2}x+\dfrac{1}{2}\)

=>\(\dfrac{1}{2}x-\sqrt{x}+\dfrac{1}{2}=0\)

=>\(x-2\sqrt{x}+1=0\)

=>x=1(loại)

c: \(\left\{{}\begin{matrix}x-\sqrt{x}+1>0\\x+1>0\end{matrix}\right.\)

=>D>0

=>D>=|D|