\(2x^2+x-\sqrt{3}=\sqrt{3}x+1\)
=>\(2x^2+x\left(1-\sqrt{3}\right)-\sqrt{3}-1=0\)
\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\cdot\left(-\sqrt{3}-1\right)\)
\(=4-2\sqrt{3}+8\sqrt{3}+8=12+6\sqrt{3}=\left(3+\sqrt{3}\right)^2>0\)
=>Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\sqrt{\left(3+\sqrt{3}\right)^2}}{2\cdot2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+\sqrt{\left(3+\sqrt{3}\right)^2}}{2\cdot2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=\dfrac{-1+\sqrt{3}-3-\sqrt{3}}{4}=\dfrac{-4}{4}=-1\\x_2=\dfrac{-1+\sqrt{3}+3+\sqrt{3}}{4}=\dfrac{2+2\sqrt{3}}{4}=\dfrac{\sqrt{3}+1}{2}\end{matrix}\right.\)
