B) (x-3)(x-7)-6x=2x+8(5x-3)²; 2 5 2

в) $$\frac{(x-3)(x-7)}{2} - 6x = \frac{2x+8}{5} - \frac{(5x-3)^2}{2}$$

Умножим обе части уравнения на 10:

$$5(x-3)(x-7) - 60x = 2(2x+8) - 5(5x-3)^2$$

$$5(x^2-7x-3x+21) - 60x = 4x+16 - 5(25x^2-30x+9)$$

$$5(x^2-10x+21) - 60x = 4x+16 - 125x^2+150x-45$$

$$5x^2-50x+105 - 60x = 4x+16 - 125x^2+150x-45$$

$$5x^2-110x+105 - 4x-16 + 125x^2-150x+45= 0$$

$$130x^2-264x+134= 0$$

$$65x^2-132x+67= 0$$

$$D = (-132)^2 - 4 \cdot 65 \cdot 67 = 17424 - 17420 = 4$$

$$x_1 = \frac{132 + \sqrt{4}}{2 \cdot 65} = \frac{132+2}{130} = \frac{134}{130} = \frac{67}{65}$$

$$x_2 = \frac{132 - \sqrt{4}}{2 \cdot 65} = \frac{132-2}{130} = \frac{130}{130} = 1$$

Ответ: $$x=\frac{67}{65}; x=1$$