Question

Решите уравнения 1.(х+2)(х-3)-3=(х+1)^2 2.(у-1)^2-(у+1)(у-7)=0

Answer 1

$1)\(x+2)(x-3)-3=(x+1)^2\x^2-3x+2x-6-3=x^2+2x+1\x^2-x^2-3x+2x-2x-6-3-1=0\-3x-10=0\-3x=10 |:(-3)\x=-10/3=-3frac{1}{3} \Otvet: -3frac{1}{3}\ \2)\(y-1)^2-(y+1)(y-7)=0\y^2-2y+1-(y^2-7y+y-7)=0\y^2-2y+1-y^2+7y-y+7=0\-2y+7y-y+1+7=0\4y=-8 |:4\y=-2\Otvet: -2$

Answer 2

Спасибо!

Answer 3

1) x^2 + 2x - 3x - 6 - 3 = x^2 + 2x + 1

3x = -10; x = -10/3

2) y^2 -2y + 1 - y^2 - y + 7y + 7 = 0

4y = -8; y = -2