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matlyubax
@matlyubax
July 2022
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решите уравнения:
1)(х+3)(х в квадрате +3х+9)-(3х-17)=х в кубе-12
2)5х-(4-2х+х в квадрате)(х+2)+х(х-1)(х-1)=0
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viktoriyamart
1) (х+3) ( x² + 3х + 9) - (3х-17) = х³ - 12
х³+3х²+9х+3х²+9х+27 - 3х + 17 - х³+ 12 = 0
х³ и -х³ взаимно уничтожаем
6х²+15х + 56 = 0
Д = b²-4ас = 15² - 4×6×56 = 225 -1344 = -1119 < 0 корней нет
2) 5х-(4-2х+х²)(х+2)+(х-1)(х-1) = 0
5х -(х³-2х²+4х+2х²-4х+8) + х²-2х+1 = 0
5х - х³ + 2х² - 4х - 2х² + 4х + 8 + х² -2х +1 = 0
-х³ +х²+3х+9 = 0
(х-3)(-х²-2х-3) = 0
х - 3 = 0 или -х²-2х-3 ≠ 0
х₁=3
-х²-2х-3 = 0 Ι ÷ (-1)
х²+2х+3 = 0
Д=b²-4ас = 2²-4×1×3 = 4 - 12 = -8 < 0 нет корней
2 votes
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Answers & Comments
х³+3х²+9х+3х²+9х+27 - 3х + 17 - х³+ 12 = 0
х³ и -х³ взаимно уничтожаем
6х²+15х + 56 = 0
Д = b²-4ас = 15² - 4×6×56 = 225 -1344 = -1119 < 0 корней нет
2) 5х-(4-2х+х²)(х+2)+(х-1)(х-1) = 0
5х -(х³-2х²+4х+2х²-4х+8) + х²-2х+1 = 0
5х - х³ + 2х² - 4х - 2х² + 4х + 8 + х² -2х +1 = 0
-х³ +х²+3х+9 = 0
(х-3)(-х²-2х-3) = 0
х - 3 = 0 или -х²-2х-3 ≠ 0
х₁=3
-х²-2х-3 = 0 Ι ÷ (-1)
х²+2х+3 = 0
Д=b²-4ас = 2²-4×1×3 = 4 - 12 = -8 < 0 нет корней