Question

Решите уравнения :
6(x+1)^2+2(x-1)(x^2+x+1)-2(x+1)^3=32
5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=15
(x+2)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=42

^2=квадрат
^3=куб

Answer 1

$6(x+1)^2+2(x-1)(x^2+x+1)-2(x+1)^3=32\ \ 6(x^2+2x+1)+2(x^3-1)-2(x^3+3x^2+3x+1)=32\ \ 6x^2+12x+6+2x^3-2-2x^3-6x^2-6x-2=32\ \ 6x=30\ \ x=30:6\ \ x=5$



$5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=15\ \ 5x(x^2-6x+9)-5(x^3-3x^2+3x-1)+15(x^2-4)=15~~|:5\ \ x^3-6x^2+9x-x^3+3x^2-3x+1+3x^2-12=3\ \ 6x=14\ \ x= frac{7}{3}$



$(x+2)^3-x(3x+1)^2+(2x+1)(4x^2-2x+1)=42\ \ x^3+6x^2+12x+8-x(9x^2+6x+1)+8x^3+1=42\ \ x^3+6x^2+12x+8-9x^3-6x^2-x+8x^3+1=42\ \ 11x=33\ \ x=3$