(х+3)*корень х-1=0
(х-1)(х+3)=0 ОДЗ {х-1>0 ,x+3>0} {x>1, x>-3} {x>1}
x^2-x+3x-3=0
x^2+2x-3=0
x2 + 2x - 3 = 0D = b2 - 4acD = 4 + 12 = 16 = 4
x1,2 = -b ± √D/2ax1 = -2 + 4/2= 2/2 = 1x2 = -2 - 4 /2= - 6/2 = -3ответ: x1 = 1; x2 = -3
(х+3)*корень х-1=0
(х-1)(х+3)=0 ОДЗ {х-1>0 ,x+3>0} {x>1, x>-3} {x>1}
x^2-x+3x-3=0
x^2+2x-3=0
x2 + 2x - 3 = 0
D = b2 - 4ac
D = 4 + 12 = 16 = 4
x1,2 = -b ± √D/2a
x1 = -2 + 4/2= 2/2 = 1
x2 = -2 - 4 /2= - 6/2 = -3
ответ: x1 = 1; x2 = -3