Question

(x^2+3x-1)^2-12x^2-36x+39=0 решите плес

Answer 1

$(x^2+3x-1)^2-12(x^2+3x-1)+27=0 \ \ t=x^2+3x-1 \ \ t^2-12t+27=0 \ \ D=144-4*27=36 \ \ t_{1} =(12-6)/2=3 \ \ t_{2} =(12+6)/2=9 \ \ 1)x^2+3x-1=3 \ x^2+3x-4=0 \ x^2+4x-x-4=0 \ x(x+4)-(x+4)=0 \ (x+4)(x-1)=0 \ x_{1} =-4; x_{2} =1 \ \ 2)x^2+3x-1=9 \ x^2+3x-10=0 \ x^2+5x-2x-10=0 \ x(x+5)-2(x+5)=0 \ (x+5)(x-2)=0 \ x_{3}=-5 ; x_{4} =2 \ \ OTVET:{-5;-4;1;2}$