Question

Pt 2x^2-3x-4=0 A= (1/x1)^2+(1/x2)^2 giúp tớ

Answer 1

\(2x^2-3x-4=0\)

\(\Delta=3^2+4.2.4=41>0\)

⇒ Phương trình có hai nghiệm phân biệt

Theo Viét : \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{3}{2}\\x_1.x_2=-2\end{matrix}\right.\)

Lại có : \(A=\left(\dfrac{1}{x_1}\right)^2+\left(\dfrac{1}{x_2}\right)^2=\dfrac{1}{x_1^2}+\dfrac{1}{x_2^2}\)\(=\dfrac{x_1^2+x_2^2}{\left(x_1x_2\right)^2}=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{\left(x_1x_2\right)^2}=\dfrac{\left(\dfrac{3}{2}\right)^2+4}{\left(-2\right)^2}=\dfrac{25}{16}\)

Vậy....