Question

Помогите алгебра! Срочно!

Answer 1

Ответ:

120. Розв'яжіть рівняння:

$1)(3x+2)(x-4)=5\\3x^{2} - 12x + 2x - 8 = 5\\3x^{2} - 10x - 13 = 0\\D = (-10)^{2} - 4 * 3 *(-13) = 100 + 156 = 256\\x_{1/2} = \frac{10б16}{6} \\x_{1} = \frac{10-16}{6} = - \frac{6}{6} = -1\\x_{2} = \frac{10+16}{6} = \frac{26}{6} =\frac{13}{3} = 4\frac{1}{3}$

$2)(x+1)(x-2)-(4x-3)(x+5)=x(x-9)\\x^{2} - 2x + x - 2 - 4x^{2} -20x +3x + 15 = x^{2} - 9x\\-3x^{2} - 18x + 13 - x^{2} + 9x = 0\\-4x^{2} - 9x + 13 = 0 (*(-1))\\4x^{2} + 9x - 13 = 0\\D = 9^{2} - 4 * 4 * (-13) = 81 + 208 = 289\\x_{1/2} = \frac{-9б17}{8}\\x_{1} = \frac{-9-17}{8} = - \frac{26}{8} = - \frac{13}{4} = -3 \frac{1}{4} = -3,25\\x_{2} = \frac{-9+17}{8} = \frac{8}{8} = 1$

$3) (5x-3)^{2} + (2x-3)(2x+3) = 2\\25x^{2} - 30x + 9 + 4x^{2} - 9 = 2\\29x^{2} - 30x -2 = 0\\D = (-30)^{2} - 4*29*(-2) = 900+ 232 = 1132\\x_{1/2} = \frac{30б\sqrt{1132} }{58} = \frac{30б\sqrt{4*283} }{59} = \frac{30б2\sqrt{283} }{59} = \frac{15б\sqrt{283} }{29}$

$4) (3x+1)(9x^{2} - 3x + 1)-(9x^{2} + 2)(3x-2) = 16x^{2} + 1\\27x^{3} - 9x^{2} + 3x + 9x^{2} - 3x + 1 -27x^{3} +18x^{2} - 6x +4 = 16x^{2} + 1 \\27x^{3} - 9x^{2} + 3x + 9x^{2} - 3x + 1 -27x^{3} +18x^{2} - 6x +4 -16x^{2} - 1 = 0\\2x^{2} - 6x + 4 = 0(:2)\\x^{2} - 3x + 2 = 0\\D = (-3)^{2} - 4 * 1 * 2 = 9 - 8 = 1\\x_{1/2} = \frac{3б1}{2}\\x_{1} = \frac{3-1}{2} = \frac{2}{2} = 1\\x_{2} = \frac{3+1}{2} = \frac{4}{2} = 2$