\(2.\)
\(a.x^2-5x+6=0\)
\(\Delta=b^2-4ac=\left(-5\right)^2-4.1.6=1>0\Rightarrow\sqrt{\Delta}=\sqrt{1}=1\)
Vậy phương trình (a) có 2 nghiệm phân biệt.
\(x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+1}{2.1}=3\)
\(x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-1}{2.1}=2\)
Vậy phương trình có \(x_1=3,x_2=2\)
\(b.x^2+3x-2=0\)
Ta có: \(\Delta=3^2-4.1.\left(-2\right)=17>0\)
ĐL Vi - et:
\(\left[{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}=\dfrac{-3}{1}=-3\\x_1.x_2=\dfrac{c}{a}=\dfrac{-2}{3}\end{matrix}\right.\)
Theo đề bài ta có:
\(\dfrac{x_1^2-x_2^2}{x_1^2x_2-x_1x_2^2}=\dfrac{x_1^2+x_2^2+2x_1x_2-2x_1x_2}{-x_1x_2\left(x_1+x_2\right)}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{-x_1x_2\left(x_1+x_2\right)}=\dfrac{\left(-3\right)^2-2\left(-\dfrac{2}{3}\right)}{\dfrac{2}{3}.\left(-3\right)}=\dfrac{-31}{6}\)
