Question

Помогите решить (рациональные уравнения)
x^3+2x^2=0
(x-6)^2+x2(x-6)=0
x^3+3x^2+5x+15=0
x^4-3x-x+3=0
x^2-24x+144=0
25x^2=60x+36=0

Answer 1

1.

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

2.

${(x - 6)}^{2} + 2x \times (x - 6) = 0 \\ {x}^{2} - 12x + 36 + 2 {x}^{2} - 12x = 0 \\ 3 {x}^{2} - 24x + 36 = 0 \\ {x}^{2} - 8x + 12 = 0 \\ d = ( - 8) ^{2} - 4 \times 1 \times 12 = 64 - 48 = 16 = {4}^{2} \\ x1 = \frac{8 + 4}{2} = \frac{12}{2} = 6 \\ x2 = \frac{8 - 4}{2} \frac{4}{2} = 2$

3.

${x}^{3} + 3 {x}^{2} + 5x + 15 = 0 \\ {x}^{2} \times (x + 3) + 5 \times (x +3) = 0 \\ (x + 3) \times ( {x}^{2} + 5) = 0 \\ x + 3 = 0 \\ {x}^{2} + 5 = 0 \\ x = - 3 \\ x = -$

4.

${x}^{4} - 3x - x + 3 = 0 \\ x \times ( {x}^{3} - 1) - 3 \times (x - 1) = 0 \\ x \times (x - 1) \times ( {x}^{2} + x + 1) - 3 \times (x - 1) = 0 \\ (x - 1) \times (x \times ( {x}^{2} + x + 1) - 3) = 0 \\ (x - 1) \times (x - 1) \times ( {x}^{2} + 2x + 3) = 0 \\ (x - 1)^{2} \times ( {x}^{2} + 2x + 3) = 0 \\ {(x - 1)}^{2} = 0 \\ {x}^{2} + 2x + 3 = 0 \\ x = 1 \\ x = -$

5.

${x}^{2} - 24x + 144 = 0 \\ (x - 12)^{2} = 0 \\ x - 12 = 0 \\ x = 12$

6 уравнение решить не могу, т.к. не знаю, как решаются уравнения с двумя знаками равенства.