Ta có: \(\left(x+1\right)\left(x+2\right)=3\)
\(\Leftrightarrow x^2+3x+2-3=0\)
\(\Leftrightarrow x^2+3x-1=0\)
\(\text{Δ}=9^2-4\cdot1\cdot\left(-1\right)=13\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{13}}{2}\\x_2=\dfrac{-3+\sqrt{13}}{2}\end{matrix}\right.\)
\(x^2+3x+2-3=0\\ \Leftrightarrow x^2+3x-1=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{-3+\sqrt{13}}{2}\\x=\dfrac{-3-\sqrt{13}}{2}\end{matrix}\right.\)