5.28. a) (3x-4)² + (2x-5) (x-1) = 1 + (x+2)²; 5 2 5

а) $$\frac{(3x-4)^2}{5} + \frac{(2x-5)(x-1)}{2} = 1 + \frac{(x+2)^2}{5}$$

Умножим обе части уравнения на 10:

$$2(3x-4)^2 + 5(2x-5)(x-1) = 10 + 2(x+2)^2$$

$$2(9x^2-24x+16) + 5(2x^2-2x-5x+5) = 10 + 2(x^2+4x+4)$$

$$18x^2-48x+32 + 10x^2-35x+25 = 10 + 2x^2+8x+8$$

$$28x^2-83x+57 - 10 - 2x^2-8x-8= 0$$

$$26x^2-91x+39= 0$$

$$D = (-91)^2 - 4 \cdot 26 \cdot 39 = 8281 - 4056 = 4225 = 65^2$$

$$x_1 = \frac{91 + \sqrt{4225}}{2 \cdot 26} = \frac{91+65}{52} = \frac{156}{52} = 3$$

$$x_2 = \frac{91 - \sqrt{4225}}{2 \cdot 26} = \frac{91-65}{52} = \frac{26}{52} = \frac{1}{2}$$

Ответ: $$x=3; x=\frac{1}{2}$$