найдите корни уравнения
(2x+1)^2=2x+1
(y-2)^2-4=0
(3x-1)^2=2(3x-1)
9-(2x-3)^2=0
(2x+1)^2=2x+1
(y-2)^2-4=0
(3x-1)^2=2(3x-1)
9-(2x-3)^2=0
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(2x+1)²=2x+1(2х+1)(2х+1-1)=0
2х(2х+1)=0
х=0 или х=-0,5
(y-2)²-4=0
(у-2-2)(у-2+2)=0
(у-4)·у=0
у=0 или у=4
(3x-1)²=2(3x-1)
(3х-1)(3х-1-2)=0
(3х-1)(3х-3)=0
3х-1=0 или 3х-3=0
х=1/3 или х=1
9-(2x-3)²=0
3²-(2х-3)²=0
(3+2х-3)(3-2х+3)=0
2х(6-2х)=0
2≠0, х=0 или 6-2х=0
2х=6
х=3