Question

Se știe ca x+y=5, iar xy=6
A. Calculați (x+y)^2
B. Arătați ca x^2 + y^2 =13
C. Calculați valoarea numărului a=(x-y)^2 +1/x + 1/y

Answer 1

x+y=5,⇒ x =5 - y

xy=6 ⇒ (5 - y) × y =6⇒ 5y - y² = 6

6+y²-5y =0

y²-5y+6=0

Δ=25-24=1

y₁=(5+1)/2=3⇒ x₁=2

y₂=(5-1)/2=2⇒ x₂=3

deci y₁ = 3⇒ x₁= 2 si y₂ = 2⇒ x₂ = 3

a) (x+y)²=(3+2)²=5²=25

b)

x²+y²=13

2²+3²=4+9=13

c)y₁ = 3⇒ x₁= 2 si y₂ = 2⇒ x₂ = 3

a=(x-y)² +1/x + 1/y =(2-3)²+1/2+1/3=1+1/2+1/3=(6+2+3)/6=11/6

a= (3-2)²+1/3+1/2=1+1/3+1/2=11/6

Answer 2

$\it x+y=5\ \ \ \ (1) \\ \\ \\ xy = 6 \ \ \ \ \ \ \ \ (2) \\ \\ \\ a)\ (x+y)^2\ \stackrel{(1)}{=} \ 5^2 = 25 \ \ \ \ \ \ (3) \\ \\ \\ b)\ x^2+y^2 = (x+y)^2 -2xy \ \stackrel{(3), (2)}{=} \ 25-2\cdot6 =25-12=13.$

$\it c)\ (x-y)^2 +\dfrac{1}{x} +\dfrac{1}{y} = (x+y)^2 -4xy ++\dfrac{x+y}{xy} = 25 - 4\cdot6+\dfrac{5}{6} = 1+\dfrac{5}{6}= \\ \\ \\ = \dfrac{11}{6}$