Câu hỏi của Triệu Nguyễn

\(S=N-M\)\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)Vì \(\left\{{}\begin{matrix}\sqrt{x}>0\\x+\sqrt{x}+1>0\end{matrix}\right.\forall x\) thỏa mãn ĐKXĐnên S>0=>|S|=SCâu trả lời: \(S=N-M\)\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)Vì \(\left\{{}\begin{matrix}\sqrt{x}>0\\x+\sqrt{x}+1>0\end{matrix}\right.\forall x\) thỏa mãn ĐKXĐnên S>0=>|S|=S