Question

решите уравнение (2х-7)(х+1)+3(4x+1)=2(5х-2)^2-53
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Answer 1

$(2x - 7)(x + 1) + 3(4x + 1) = 2(5x - 2)^{2} - 53$

$2x^{2} + 2x - 7x - 7 + 12x + 3 = 2(25x^{2} - 20x + 4) - 53$

$2x^{2} + 7x - 4 = 50x^{2} - 40 x + 8 - 53$

$2x^{2} + 7x -4 = 50x^{2} - 40x - 45$

$48x^{2} - 47x = 41 \ \ \ | : 48$

$x^{2} - \dfrac{47}{48}x = \dfrac{41}{48}$

$x^{2} - \dfrac{47}{48}x + \left(\dfrac{47}{96} \right)^{2} = \dfrac{41}{48} + \left(\dfrac{47}{96} \right)^{2}$

$\left(x - \dfrac{47}{96} \right)^{2} = \dfrac{10081}{9216}$

Первый случай:

$x_{1} - \dfrac{47}{96} = -\sqrt{\dfrac{10081}{9216} }\\x_{1} =\dfrac{47}{96} - \dfrac{\sqrt{10081}}{96} \\x_{1} = \dfrac{47 - \sqrt{10081}}{96}$

Второй случай:

$x_{2} - \dfrac{47}{96} = \sqrt{\dfrac{10081}{9216} }\\x_{2} =\dfrac{47}{96} + \dfrac{\sqrt{10081}}{96} \\x_{2} = \dfrac{47 + \sqrt{10081}}{96}$

Ответ: $x_{1} = \dfrac{47 - \sqrt{10081}}{96}; \ x_{2} = \dfrac{47 + \sqrt{10081}}{96}$