126. Решите уравнение: 1) (2x+5)(x + 2) = 21; 2) (x + 3)(x-1)-(3x+1)(x - 7) = x(x + 18); 3) (4x-3)² + (2x-1)(2x+1) = 24.

Решим уравнения:

1. $$(2x+5)(x+2)=21$$

   $$2x^2+4x+5x+10-21=0$$

   $$2x^2+9x-11=0$$

   $$D = b^2 - 4ac = (9)^2 - 4 \cdot 2 \cdot (-11) = 81 + 88 = 169$$

   $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{-9 + \sqrt{169}}{2 \cdot 2} = \frac{-9 + 13}{4} = \frac{4}{4} = 1$$

   $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{-9 - \sqrt{169}}{2 \cdot 2} = \frac{-9 - 13}{4} = \frac{-22}{4} = -5,5$$

   Ответ: -5,5; 1

2. $$(x+3)(x-1)-(3x+1)(x-7)=x(x+18)$$

   $$x^2-x+3x-3-(3x^2-21x+x-7)=x^2+18x$$

   $$x^2+2x-3-3x^2+20x+7=x^2+18x$$

   $$-2x^2+22x+4=x^2+18x$$

   $$3x^2-4x-4=0$$

   $$D = b^2 - 4ac = (-4)^2 - 4 \cdot 3 \cdot (-4) = 16 + 48 = 64$$

   $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{4 + \sqrt{64}}{2 \cdot 3} = \frac{4 + 8}{6} = \frac{12}{6} = 2$$

   $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{4 - \sqrt{64}}{2 \cdot 3} = \frac{4 - 8}{6} = \frac{-4}{6} = -\frac{2}{3}$$

   Ответ: $$\frac{-2}{3}; 2$$

3. $$(4x-3)^2+(2x-1)(2x+1)=24$$

   $$16x^2-24x+9+4x^2-1=24$$

   $$20x^2-24x-16=0$$

   $$5x^2-6x-4=0$$

   $$D = b^2 - 4ac = (-6)^2 - 4 \cdot 5 \cdot (-4) = 36 + 80 = 116$$

   $$x_1 = \frac{-b + \sqrt{D}}{2a} = \frac{6 + \sqrt{116}}{2 \cdot 5} = \frac{6 + 2\sqrt{29}}{10} = \frac{3 + \sqrt{29}}{5}$$

   $$x_2 = \frac{-b - \sqrt{D}}{2a} = \frac{6 - \sqrt{116}}{2 \cdot 5} = \frac{6 - 2\sqrt{29}}{10} = \frac{3 - \sqrt{29}}{5}$$

   Ответ: $$\frac{3 + \sqrt{29}}{5}; \frac{3 - \sqrt{29}}{5}$$