Question

8+x²+6x=0 , (x-1)³+(3+2x³)=3(3x³+4x²-2x-3) va rog dau coroana

Answer 1

[tex]x^2+6x+9-1=0\\ (x+3)^2-1=0\\(x+3-1)(x+3+1)=0\\(x+2)(x+4)=0\\ x+2=0,x=-2\\ x+4=0,x=-4[/tex]
La a doua ecuatie, nu cumva este (3+2x)^3?

In aceasta situatie solutia este:

[tex]x^3-3x^2+3x-1+27+54x+36x^2+8x^3=9x^3+12x^2-6x-9\\ 33x^2+57x+26=12x^2-6x-9\\ 21x^2+63x+35=0|:7\\ 3x^2+9x+5=0\\ \triangle=b^2-4ac=81-60=21\\ x_1=\frac{-9+\sqrt{21}}{6}\\ x_2=\frac{-9-\sqrt{21}}{6} [/tex]