Question

Proszę o jak najszybszą pomoc :)
Rozwiąż równiania :

a)x3-4x2+5x-20=0
b)x3-5x2-2x+10=0
c)x3+5x2-x-5=0
d)2x3+3x2-4x-6=0
e)x3+3x2-9x-27=0
f)x3-3x3+4x-12=0
g)x3-3x2-4x+12=0
h)x3-2x2-9x+18=0
i)2x3-x2-8x+4=0
j)6x3+2x2-3x-1=0

Answer 1

A) x(x^2+5) - 4(x^2 + 5) = 0
(x^2 + 5)*(x-4) = 0
x^2 = -5 =>sprzeczne bo cokolwiek podniesione do kwadratu daje liczbę dodatnią
x - 4 = 0 =>x = 4 
b) x(x^2 - 2) - 5(x^2 -2) = 0
(x^2 - 2)*(x-5) = 0
x^2 -2 = 0 => x^2 = 2
x1 = sqrt(2)
x2 = -sqrt(2)
x-5 = 0=> x3 = 5

c) x(x^2 - 1) +5(x^2 - 1) =0
(x^2 - 1) * (x + 5) =0
x^2 - 1 = 0 => x^2= 1
x1 = 1
x2 = -1
x+5 = 0=> x3 = -5

d) 2x(x^2 - 2) + 3(x^2 - 2) = 0
(x^2 - 2)*(2x + 3) =0
x^2 - 2 = 0 => x^2 = 2
x1 = sqrt(2)
x2 = -sqrt(2)
2x + 3 = 0 => 2x = - 3//:2 => x3 = -3/2

e) x(x^2 -9) + 3(x^2 -9)=0
(x^2 -9)*(x +3) = 0
x^2 - 9 = 0 => x^2 = 9
x1 = 3
x2 = -3 => podwójne bo x+3 = 0 => x = -3

f) x(x^2 +4) -3(x^2 +4) = 0
(x^2 + 4)*(x - 3) = 0
x-3 = 0 => x1 = 3
x^2 = -4 => sprzeczne bo 2 potęga nie może być ujemna

g) x(x^2 - 4) -3(x^2 -4) = 0
(x^2 - 4)*(x - 3) = 0
x-3=0 => x1 = 3
x^2 -4 =0 => x^2 =4 => x2=2 x3 = -2

h) x(x^2 -9) -2(x^2 - 9)=0
(x^2 -9)*(x-2) = 0
x-2 = 0 => x1 = 2
x^2 - 9= 0=> x^2 =9=> x2 = 3 x3 = -3

i) 2x(x^2 - 4) - (x^2 -4) = 0
(x^2 -4)*(2x - 1) =0
2x - 1 = 0=> 2x = 1 => x1 = 1/2
x^2 - 4 =0 => x^2 = 4 => x2 = 2 x3 = -2

j) 3x(2x^2 - 1) + (2x^2 - 1) = 0
(2x^2 -1)*(3x + 1) = 0
3x + 1 = 0 => 3x = -1 => x1 = -1/3
2x^2 - 1 = 0 => 2x^2 = 1 = > x^2 = 1/2 => x2 = sqrt(2)/2 x3 = -sqrt(2)/2