Question

Biết \(4x^2-2x-1=0\) có 2 nghiệm `x_1 ,x_2`. Tính giá trị \(A=\left(x_1-x_2\right)^2-x_1\left(x_1-\dfrac{1}{2}\right)\)

Answer 1

Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=\dfrac{-\left(-2\right)}{4}=\dfrac{1}{2}\\x_1\cdot x_2=\dfrac{c}{a}=\dfrac{-1}{4}\end{matrix}\right.\)

\(A=\left(x_1-x_2\right)^2-x_1\left(x_1-\dfrac{1}{2}\right)\)

\(=\left(x_1+x_2\right)^2-4x_1x_2-x_1^2+\dfrac{1}{2}x_1\)

\(=\left(x_1+x_2\right)^2-4x_1x_2-x_1^2+x_1\left(x_1+x_2\right)\)

\(=\left(x_1+x_2\right)^2-4x_1x_2+x_1x_2\)

\(=\left(x_1+x_2\right)^2-3x_1x_2\)

\(=\left(\dfrac{1}{2}\right)^2-3\cdot\dfrac{-1}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)