Решите (2x+1)(x+√2)=0 2x-3=1-5x (2x-3)(2x+3)=7 (2x-5)(x+3)=0

(2х+1)(х+√2)=0
2х+1=0, х=-1/2
х+√2=0, х=-√2

2х-3=1-5х
7х=4
х=4/7

(2х-3)(2х+3)=7
4х^2-9=7
4х^2=16
х^2=4
х=2,х=-2

(2х-5)(х+3)=0
2х-5=0, х=2,5
х+3=0, х=-3