Question

(2x-3)^2+(x+1)^2=(3x-2)^2-4(x+7)^2-140x+4
ПОМОГИТЕ

Answer 1

(2х-3)^2+(х+1)^2=(3х-2)^2-4(х+7)^2-140х+4
(4х^2-12х+9)+(х^2+2х+1)=(9х^2-12х+4)-4(х^2+14х+49)-140х+4
4х^2-12х+9+х^2+2х+1=9х^2-12х+4-4х^2-56х-196-140х+4
4х^2+х^2-9х^2+4х^2-12х+2х+12х+56х+140х=4-196+4-9-1
198х=-198
х=-1
Ответ: х=-1

Answer 2

$(2x-3)^2+(x+1)^2=(3x-2)^2-4(x+7)^2-140x+4\4x^2-12x+9+(x+1)^2=(3x-2)^2-(2(x+7))^2-140x+4\4x^2+128x+(x+1)^2=(3x-2-2(x+7))(3x-2+2(x+7))-5\4x^2+128x+(x+1)^2=(3x-2-2x-14)(3x-2+2x+14)-5\4x^2+128x+x^2+2x+1=(x-16)(5x+12)-5\5x^2+130x=5x^2+12x-80x-192-6\198x=-198\x=-1$