Question

$(x^{2}+2x)^{2}-2(x+2)(x^{2}-x)-15(x^{2}-2x+1)=0$
Найдите корни уравнения

Answer 1

$(x^2+2x)^2-2(x+2)(x^2-x)-15(x^2-2x+1)=0, \ (x(x+2))^2-2x(x+2)(x-1)-15(x-1)^2=0, \ ((x(x+2))^2-2x(x+2)(x-1)+(x-1)^2)-16(x-1)^2=0, \ (x(x+2)-(x-1))^2=16(x-1)^2, \ (x^2+x+1)^2=(4x-4)^2, \ left [ {{x^2+x+1=4x-4,} atop {x^2+x+1=-(4x-4);}} right. \ x^2-3x+5=0, \ D=-11<0; \ x^2+5x-3=0, \ D=37, \ x_1=-2,5-0,5sqrt{37}, \ x_2=-2,5+0,5sqrt{37}.$