Question

(2х-1)в квадрате -7(2х-1)+4=0

Answer 1

$(2x - 1) {}^{2} - 7(2x - 1) + 4 = 0 \\ 2x - 1 = a \\ {a}^{2} - 7a + 4 = 0 \\ D = ( - 7 {)}^{2} - 4 \times 4 = 49 - 16 = 33 \\ a_{1} = \frac{7 - \sqrt{33} }{2} \\ a_{2} = \frac{7 + \sqrt{33} }{2} \\ \\ 1) \: a = \frac{7 - \sqrt{33} }{2} \\ 2x - 1 = \frac{7 - \sqrt{33} }{2} \\ 2x = \frac{7 - \sqrt{33} }{2} + 1 \\ 2x = \frac{7 - \sqrt{33} + 2 }{2} \\ 2x = \frac{9 - \sqrt{33} }{2} \\ x = \frac{9 - \sqrt{33} }{2} \div 2 \\ x_{1} = \frac{9 - \sqrt{33} }{4} \\ \\ 2) \: a = \frac{7 + \sqrt{33} }{2} \\ 2x - 1 = \frac{7 + \sqrt{33} }{2} \\ 2x = \frac{7 + \sqrt{33} }{2} + 1 \\ 2x = \frac{7 + \sqrt{33} + 2 }{2} \\ 2x = \frac{9 + \sqrt{33} }{2} \\ x = \frac{9 + \sqrt{33} }{2} \div 2 \\ x_{2} = \frac{9 + \sqrt{33} }{4}$

Answer 2

Ответ:

(9+√33)/4;  (9-√33)/4

Объяснение: