\(\sqrt{x}+\sqrt{x+1}-\sqrt{x^2+x}=1\) ĐKXĐ: x ≥ 0
⇔ \(\sqrt{x}-\sqrt{x\left(x+1\right)}+\sqrt{x+1}-1\) = 0
⇔ \(-\sqrt{x}\left(\sqrt{x+1}-1\right)+\left(\sqrt{x+1}-1\right)=0\)
⇔ \(\left(\sqrt{x+1}-1\right)\left(1-\sqrt{x}\right)=0\)
⇔ \(\left[{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{x}=1\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x+1=1\\x=1\end{matrix}\right.\)⇔ \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy nghiệm của pt là x = 0; x = 1