Question

(x+2)^2/2-x^2-4/4-(x-2)^2/8=x^2/8
очень срочно, плиииззз!!!!

Answer 1

((x+2)^2/2)-((x^2-4)/4)-((x-2)^2/8)=x^2/8, (4(x+2)^2/8)-(2(x^2-4)/8)-((x-2)^2/8)-(x^2/8)=0, (4(x^2+4x+4)-2(x^2-4)-(x^2-4x+4)-x^2)/8=0 (4x^2+16x+16-2x^2+8-x^2+4x-4-x^2)/8=0, (20x+20)/8=0, 20x+20=0, x=-1.