Найдите корни уравнения:[tex]( x^{2} - 1)^{3} + 2( x^{2} - 1) ^{2} = 0 x^{2} (x-1)-3x(x-1)=0[/tex]. Created by 5162. algebra-ru

tex]...

July 2022 1 9 Report

Найдите корни уравнения:
[tex]( x^{2} - 1)^{3} + 2( x^{2} - 1) ^{2} = 0

x^{2} (x-1)-3x(x-1)=0[/tex]

Answers & Comments

Ртуть3 1) (x² - 1)³ + 2(x² - 1)² = 0
$x^{6}-3x^{4}+3x^{2}-1+2(x^{4}-2x^{2}+1)=0
$
$x^{6}-3x^{4}+3x^{2}-1+2x^{4}-4x^{2}+2=0$
$x^{6}-x^{4}-x^{2}+1=0$
$x^{4}(x^{2}-1)-(x^{2}-1)=0
$
$(x^{4}-1)(x^{2}-1)=0$
$(x^{2}-1)(x^{2}+1)(x^{2}-1)=0$
$(x^{2}-1)^{2}(x^{2}-1)=0$
(x² - 1)² = 0 или x² + 1 = 0
x² - 1 = 0 x² = -1
x² = 1 ∅
x = +- 1
Ответ: x = -1
x = 1

2) x²(x - 1) - 3x(x - 1) = 0
(x - 1)(x² - 3x) = 0
x² - 3x = 0 или x - 1 = 0
x(x - 3)=0 x = 1
x = 0 x - 3 = 0
x = 3
Ответ: x = 0
x = 1
x = 3

1 votes Thanks 1

Answers & Comments

Helpful Links

Helpful Social