Question

Решите уравнение 1) (x^2-7x)+x-7=0 2)3x^2-9x+(x-3)=0 3)(×^2-6x)-4x+24=0 4)7x^2+28x-(x+4)=0
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Answer 1

4 не правильно решил, я хз как делать 4

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Answer 2

Ответ:

1)

$x1 = - 1 \\ x2 = 7$

2)

$x1 = - \frac{1}{3} \\ x2 = 3$

3)

$x1 = 4 \\ x2 = 6$

4)

$x1 = - 4 \\ x2 = \frac{1}{7}$

Пошаговое объяснение:

1)

$(x {}^{2} - 7x) + x - 7 = 0 \\ x {}^{2} - 7x + x - 7 = 0 \\x \times (x + 1) - 7 \times (x + 1) = 0 \\(x + 1) \times (x - 7) = 0 \\ x + 1 = 0 \\ x - 7 = 0 \\ x1 = - 1 \\ x2 = 7$

2)

$3x {}^{2} - 9x + (x - 3) = 0 \\ 3x {}^{2} - 9x + x - 3 = 0 \\ x \times (3x + 1) - 3 \times (3x - 1) = 0 \\(3x + 1) \times (x - 3) = 0 \\ 3x + 1 = 0 \\ x - 3 = 0 \\ x1 = - \frac{1}{3} \\ x2 = 3$

3)

$(x {}^{2} - 6x) - 4x + 24 = 0 \\ x {}^{2} - 6x - 4x + 24 = 0 \\ x \times (x - 4) - 6 (x - 4) = 0 \\ (x - 4) \times (x - 6) = 0 \\ x - 4 = 0 \\ x - 6 = 0 \\ x1 = 4 \\ x2 = 6$

4)

$7x {}^{2} + 28x - (x + 4) = 0 \\ 7x {}^{2} + 28x - x - 4 = 0 \\ 7x \times (x + 4) - (x + 4) = 0 \\ (x + 4) \times (7x - 1) = 0 \\ x + 4 = 0 \\ 7x - 1 = 0 \\ x1 = - 4 \\ x2 = \frac{1}{7}$