Question

Cho pt \(x^2-3x+1=0\).

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

Answer 1

x² - 3x + 1 = 0

\(\Delta\) = (-3)² - 4.1.1 = 9 - 4 = 5 > 0

\(\Rightarrow\) x₁ = \(\dfrac{3+\sqrt{5}}{2}\); x₂ = \(\dfrac{3-\sqrt{5}}{2}\)

G = x₁(2x₁ - 3) + x₂² = \(\left(\dfrac{3+\sqrt{5}}{2}\right)\left(2\cdot\left(\dfrac{3+\sqrt{5}}{2}\right)-3\right)+\dfrac{3-\sqrt{5}}{2}\)

= \(2\cdot\dfrac{\left(3+\sqrt{5}\right)^2}{4}-\dfrac{3\left(3+\sqrt{5}\right)}{2}+\dfrac{3-\sqrt{5}}{2}\)

= \(\dfrac{9+6\sqrt{5}+5-9-3\sqrt{5}+3-\sqrt{5}}{2}\)

= \(\dfrac{8+2\sqrt{5}}{2}\) = \(4+\sqrt{5}\)

Vậy G = \(4+\sqrt{5}\)

Chúc bn học tốt!