Решите уравнение (х+1)^2=16 (x-1)^2=0 (x-5)^2=1 (2x-1)^2=4 - вопрос №31888971216120521212436210032 от ирина1857 02.09.2021 19:45

Question

Алгебра: Решите уравнение (х+1)^2=16 (x-1)^2=0 (x-5)^2=1 (2x-1)^2=4 (3x+6)^2=100 (3-2x)^2=25

ирина1857 · Answer

(х+1)^2=16     х + 1 = +-4       х₁ = 4 -1 = 3        х₂ = -4 - 1 = -5
(x-1)^2=0   х - 1 =0    х = 1
(x-5)^2=1 х -5 = +-1 х ₁= 1+5 = 6      х₂ = -1 + 5 = 4
(2x-1)^2=4    2х - 1 = +-2     2х₁ = 2+1 = 3    х₁ = 3/2
2х₂ = -2 + 1 = -1    х₂ = -1/2
(3x+6)^2=100 3х + 6 = +-10    3х₁ = 10 - 6 = 4   х₁ = 4/3
3х₂ = -10-6 = -16 х₂ = -16/3
(3-2x)^2=25 3 - 2х = +-5 2х₁ = 3 - 5 = -2 х₁ = -2/2 = -1
2х₂ = 3 + 5 = 8 х₂ = 8/2 = 4