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Oblicz sumę odwrotności kwadratów pierwiastków równania.
A) - x2+10x+10=0
B) 1/2x2+x-2=0
C) - x2+3x+7=0
Jest to zad 7 strona 29​


Odpowiedź :

Odpowiedź:

Szczegółowe wyjaśnienie:

a) -x² + 10x + 10 =0

x₁ + x₂ = [tex]\frac{-b}{a}[/tex] = [tex]\frac{-10}{-1}[/tex] = 10

x₁ * x₂ = [tex]\frac{c}{a}[/tex] = [tex]\frac{10}{-1}[/tex] = - 10

[tex]\frac{1}{(x_{1})^{2} }[/tex] +  [tex]\frac{1}{(x_{2})^{2} }[/tex] = [tex]\frac{x_{2} }{(x_{1}* x_{2})^{2} }[/tex] + [tex]\frac{x_{1} }{(x_{2}* x_{1})^{2} }[/tex] =  [tex]\frac{x_{1} +x_{2} }{(x_{2}* x_{1})^{2} }[/tex] = [tex]\frac{10}{(-10)^{2} }[/tex] =  [tex]\frac{10}{100}[/tex] = [tex]\frac{1}{10}[/tex]

b) [tex]\frac{1}{2} x^{2}[/tex] + x - 2 = 0

x₁ + x₂ = [tex]\frac{-1}{\frac{1}{2} }[/tex] = -1 * 2 = -2

x₁ * x₂ = [tex]\frac{-2}{\frac{1}{2} }[/tex] = -2 * 2 = - 4

[tex]\frac{1}{(x_{1})^{2} }[/tex] + [tex]\frac{1}{(x_{2})^{2} }[/tex] = [tex]\frac{x_{1} +x_{2} }{(x_{2}* x_{1})^{2} }[/tex] = [tex]\frac{-2}{(-4)^{2} }[/tex] = [tex]\frac{-2}{16}[/tex] = [tex]-\frac{1}{8}[/tex]

c) -x² + 3x + 7 = 0

x₁ + x₂  = [tex]\frac{-3}{-1}[/tex] = 3

x₁ * x₂ = [tex]\frac{7}{-1}[/tex] = - 7

[tex]\frac{1}{(x_{1})^{2} }[/tex] + [tex]\frac{1}{(x_{2})^{2} }[/tex] = [tex]\frac{x_{1} +x_{2} }{(x_{2}* x_{1})^{2} }[/tex] = [tex]\frac{3}{(-7)^{2} }[/tex] = [tex]\frac{3}{49}[/tex]