Question

X(x во 2степени +2+1)=2(x+1)

Answer 1

$x \times ( {x}^{2} + 2 + 1) = 2(x + 1) \\ x \times ( {x}^{2} + 3) = 2x + 2 \\ {x}^{3} + 3x = 2x + 2 \\ {x}^{3} + 3x - 2x - 2 = 0 \\ {x}^{3} + x - 2 = 0 \\ {x}^{3} - x + 2x - 2 = 0 \\ x \times ( {x}^{2} - 1) + 2(x - 1) = 0 \\ x \times (x - 1) \times (x + 1) + 2(x - 1) = 0 \\ (x - 1) \times (x \times (x + 1) + 2) = 0 \\ (x - 1) \times ( {x}^{2} + x + 2 ) = 0 \\ x - 1 = 0 \\ {x}^{2} + x + 2 = 0 \\ x = 1$

Answer 2

X(x^2+2+1)=2(x+1)
x^3+2x+x=2x+2
x^3+3x-2x-2=0
x^3+x-2=0
x^3-x+2x-2=0
x*(x^2-1)+2*(x-1)=0
x*(x-1)*(x+1)+2(x-1)=0
(x-1)*(x*(x+1)+2)=0
(x-1)*(x^2+x+2)=0
значит либо:
x-1=0 либо
x^2+x+2=0;
x=1
x€R
Ответ x=1