Question

X²+2x+1=1
9-x²=0
(x-3)(x+2)=0
решите

Answer 1

Объяснение:

1)х²+2х+1=1

х²+2х=0

х(х+2)=0

х1=0 х+2=0

х2=-2

2)9-х²=0

-х²=-9¦*-1

х²=9

х1=3 х2=-3

3)(x-3)(x+2)=0

х²+2х-3х-6=0

х²-х=6

х(х-1)=6

х1=6 х-1=6

х2=7

Answer 2

$1.x {}^{2} + 2x + 1 = 1 \\ x {}^{2} + 2x = 0 \\ x(x + 2) = 0 \\ x = 0 \: \: \: : \: \: \: x = - 2$

$2.9 - x { }^{2} = 0 \\ x {}^{2} = - 9 \\ x =+- 3 \\$

$3.(x - 3)(x + 2) = 0 \\ x {}^{2} + 2x - 3x - 6 = 0 \\ x {}^{2} - x - 6 = 0$

по теореме Виета:

$x_1 + x_2 = 1 \\ x_1 \times x_{2} = - 6 \\ x_1 = 3 \\ x_2 = - 2$