Question

(x^2+2x)^2+13×(x+1)^2-1=0

^ - степень

Answer 1

$(x^2+2x)^2+13*(x+1)^2-1=0 \(x^2+2x)^2+13(x^2+2x+1)-1=0 \y=x^2+2x \y^2+13(y+1)-1=0 \y^2+13y+13-1=0 \y^2+13y+12=0 \D=169-48=121=11^2 \y_1= frac{-13+11}{2} =-1 \y_2= frac{-13-11}{2} =-12 \x^2+2x=-1 \x^2+2x+1=0 \(x+1)^2=0 \x+1=0 \x=-1 \x^2+2x=-12 \x^2+2x+12=0 \D=4-4*12 textless 0 Rightarrow x in varnothing$
Ответ: x=-1