Question

Решите уравнения 1) 5x+(10+3x)(x-1)=(3x-1)(3x+1)-6x^2 2) x(x-1)(x+1)-x^2+2x=x^3-(x+2)^2 3) x^3+2x-1+(x-2)^2+(x+1)^3=2(2x^2+x^3)

Answer 1

1) 5x+10x-10+3x^2-3x=9x^2-1-6x^2
3x^2+12x-10=3x^2-1
12x=10-1
12x=9
4x=3
x=3/4=0,75
2) x(x^2-1)-x^2+2x=x^3-(x^2+4x+4)
x^3-x-x^2+2x=x^3-x^2-4x-4
x-x^2=-x^2-4x-4
x=-4x-4
5x=-4
x=-4/5=-0,8
3) x^3+2x-1+x^2-4x+4+x^3+3x^2+3x+1=4x^2+2x^3
2x^3+4x^2+x+4=4x^2+2x^3
x+4=0
x=-4

Answer 2

$1)5x+\left(10+3x\right)\left(x-1\right)=\left(3x-1\right)\left(3x+1\right)-6x^2 \\$
$3x^2+12x-10=3x^2-1 \\ 3x^2+12x-10+10=3x^2-1+10 \\ 3x^2+12x=3x^2+9 \\ 3x^2+12x-3x^2=3x^2+9-3x^2 \\ 12x=9 \\ x= \frac{9}{12} = \frac{3}{4} = 0,75$
$2)x\left(x-1\right)\left(x+1\right)-x^2+2x=x^3-\left(x+2\right)^2 \\ x^3-x^2+x=x^3-x^2-4x-4 \\ x^3-x^2+x-\left(x^3-x^2-4x\right)=x^3-x^2-4x-4-\left(x^3-x^2-4x\right) \\ 5x=-4 \\ x=-\frac{4}{5}=-0.8$
$3)x^3+2x-1+\left(x-2\right)^2+\left(x+1\right)^3=2\left(2x^2+x^3\right) \\ 2x^3+4x^2+x+4=4x^2+2x^3 \\ 2x^3+4x^2+x+4-4=4x^2+2x^3-4 \\ 2x^3+4x^2+x=4x^2+2x^3-4 \\ 2x^3+4x^2+x-\left(4x^2+2x^3\right)=4x^2+2x^3-4-\left(4x^2+2x^3\right) \\ x=-4$