a: \(x^2-3x-2=0\)
\(\text{Δ}=\left(-3\right)^2-4\cdot1\cdot\left(-2\right)=9+8=17>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3-\sqrt{17}}{2\cdot1}=\dfrac{3-\sqrt{17}}{2}\\x_2=\dfrac{3+\sqrt{17}}{2}\end{matrix}\right.\)
b: \(4x^2-11x+7=0\)
=>\(4x^2-4x-7x+7=0\)
=>(x-1)(4x-7)=0
=>\(\left[{}\begin{matrix}x-1=0\\4x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=\dfrac{7}{4}\end{matrix}\right.\)
c: \(-2x^2-5x+7=0\)
=>\(2x^2+5x-7=0\)
=>(2x+7)(x-1)=0
=>\(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=1\end{matrix}\right.\)
d: \(x^2-\left(2+\sqrt{3}\right)x+1+\sqrt{3}=0\)
=>\(x^2-\left(1+1+\sqrt{3}\right)x+1+\sqrt{3}=0\)
=>\(x^2-x-\left(1+\sqrt{3}\right)x+1+\sqrt{3}=0\)
=>\(x\left(x-1\right)-\left(1+\sqrt{3}\right)\left(x-1\right)=0\)
=>\(\left(x-1\right)\left[x-\left(1+\sqrt{3}\right)\right]=0\)
=>\(\left[{}\begin{matrix}x-1=0\\x-\left(1+\sqrt{3}\right)=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\\x=1+\sqrt{3}\end{matrix}\right.\)
