Cho phương trình x² - 10x + 3 = 0 có 2 nghiệm x1; x2

x² - 10x + 3 = 0

∆' = (-5)² - 3 = 22

x₁ = 5 + √22; x₂ = 5 - √22

A = x1^2 + x2^2 = (5 + √22)² + (5 - √22)²

= 25 + 10√22 + 22 + 25 - 10√22 +22 = 94

B=x1^3 + x2^3

= (5 + √22)³ + (5 - √22)³

= 125 + 3.25.√22 + 3.5.22 + (√22)³ + 125 - 3.25.√22 + 3.5.22 - (√22)³

= 910

x² - 10x + 3 = 0

∆' = (-5)² - 3 = 22

x₁ = 5 + √22; x₂ = 5 - √22

C = x1√x2 + x2√x1

= (5 + √22).√(5 - √22) + (5 - √22).√(5+√22)

= $\sqrt{5+\sqrt{22}}.\sqrt{(5+\sqrt{22})(5-\sqrt{22})}+\sqrt{5-\sqrt{22}}.\sqrt{(5+\sqrt{22})(5-\sqrt{22})}$

= $\sqrt{5+\sqrt{22}}.3+\sqrt{5-\sqrt{22}}.3$

= $3.(\sqrt{5+\sqrt{22}}+\sqrt{5-\sqrt{22}})$

D = |x1 - x2|

= |5 + √22 - (5 - √22)|

= |5 + √22 -5 + √22|

= 2√22