Odpowiedź:
[tex]\frac{2 x^2 - x - 1}{x^3 - 5 x^2 +2 x + 2} = 0[/tex] ⇔ 2 x² - x - 1 = 0 i x³ - 5 x² +2 x + 2 ≠ 0
2 x² - x - 1 = 0
Δ = ( - 1)² - 4*2*(- 1) = 1 + 8 = 9 √Δ = 3
x = [tex]\frac{1 - 3}{2*2} = \frac{- 2}{4} = - \frac{1}{2}[/tex] lub x = [tex]\frac{1 + 3}{4} = 1[/tex]
spr. 1³ - 5* 1² + 2*1 + 2 = 1 - 5 + 4 = 0 - odpada
( - [tex]\frac{1}{2}[/tex] )³ - 5*[tex]\frac{1}{4} + 2*( -\frac{1}{2}) + 2[/tex] = [tex]- \frac{1}{8} - \frac{5}{4} - 1 + 2 = - \frac{11}{8} + 1[/tex] ≠ 0
Odp. x = - [tex]\frac{1}{2}[/tex]
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Odpowiedź:
[tex]\frac{2 x^2 - x - 1}{x^3 - 5 x^2 +2 x + 2} = 0[/tex] ⇔ 2 x² - x - 1 = 0 i x³ - 5 x² +2 x + 2 ≠ 0
2 x² - x - 1 = 0
Δ = ( - 1)² - 4*2*(- 1) = 1 + 8 = 9 √Δ = 3
x = [tex]\frac{1 - 3}{2*2} = \frac{- 2}{4} = - \frac{1}{2}[/tex] lub x = [tex]\frac{1 + 3}{4} = 1[/tex]
spr. 1³ - 5* 1² + 2*1 + 2 = 1 - 5 + 4 = 0 - odpada
( - [tex]\frac{1}{2}[/tex] )³ - 5*[tex]\frac{1}{4} + 2*( -\frac{1}{2}) + 2[/tex] = [tex]- \frac{1}{8} - \frac{5}{4} - 1 + 2 = - \frac{11}{8} + 1[/tex] ≠ 0
Odp. x = - [tex]\frac{1}{2}[/tex]
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Szczegółowe wyjaśnienie: