Question

Помогите решить задание 1​

Answer 1

Ответ:

1а)

$(x-7)^2 = 7x + 149\\\\x^2 - 14x + 49 - 7x - 149 = 0\\\\x^2 - 21 x - 100 = 0\\\\D = 441 + 400 = 841\\\\\sqrt{D} = 29\\\\x_1 = \frac{21-29}{2} = \frac{-8}{2} = -4\\\\x_2 = \frac{21+29}{2} = \frac{50}{2} = 25$

1б)

$(x-2)^2 = 88-9x\\\\x^2-4x+4 - 88 + 9x = 0\\\\x^2 + 5x- 84 = 0\\\\D = 25 + 336 = 361\\\\\sqrt D = 19\\\\x_1 = \frac{-5-19}{2} = \frac{-24}{2} = -12\\\\x_2 = \frac{-5+19}{2} = \frac{14}{2} = 7$

1в)

$6(x+2)^2 = - 9x - 18\\\\6(x+2)^2 = -9(x+2) \\\\ (x+2)^2 = -1,5(x+2)\\\\ x^2 + 4x + 4 - 1,5x - 3 = 0\\\\x^2 + 2,5x + 1 = 0|\cdot 2\\\\ 2x^2 + 5x + 2 = 0\\\\ D = 25 - 16 = 9\\\\\sqrt D = 3\\\\ x_1 = \frac{-5-3}{4} = -2\\\\x_2 = \frac{-5+3}{4} = \frac{-2}{4} = -\frac{1}{2}$

1г)

$(x+2)^2 - 5 = (3-5x)^2\\\\ x^2 + 4x + 4 - 5 - 9 +30x - 25x^2 = 0\\\\-24x^2 + 34x - 10 = 0|:(-2)\\\\12x^2 - 17x + 5 = 0|\\\\D = 289 - 240 = 49\\\\\sqrt D = 7\\\\x_1 = \frac{17-7}{24} = \frac{10}{24} = \frac{5}{12}\\\\x_2 = \frac{17+7}{24} = \frac{24}{24} = 1$

2а)

$(x-5)(x+5) = 7x + 73\\\\x^2 - 25 - 7x - 73 = 0\\\\x^2 - 7x - 98\\\\D = 49 + 392 = 441\\\\\sqrt D = 21\\\\x_1 = \frac{7-21}{2} = \frac{-14}{2} = -7\\\\x_2 = \frac{7 +21}{2}= \frac{28}{2} = 14$

2б)

$(-x -2)(x-5) = x(3x-3)\\\\-x^2 + 5x - 2x + 10 = 3x^2 - 3x\\\\-x^2 - 3x^2 + 5x - 2x + 3x + 10 = 0\\\\-4x^2 + 6x + 10 = 0| :(-2)\\\\ 2x^2 - 3x - 5 = 0\\\\D = 9 + 40 = 49\\\\\sqrt D = 7\\\\x_1 = \frac{3-7}{4} = -\frac{4}{4} = -1\\\\x_2 = \frac{3+7}{4} = \frac{10}{4} = 2,5$

2в)

$-x(\frac{1}{5} - x) = (x-3)(x+3)\\\\-0,2x + x^2 = x^2 - 9\\\\x^2 - x^2 - 0,2x = -9\\\\-0,2x = -9\\\\x = 45$

2г)

$16(x-7) = (6x+4)(x-7)\\\\16x - 112 = 6x^2 - 42x + 4x - 28\\\\6x^2 - 42x+4x - 16x - 28 + 112 = 0\\\\6x^2 - 54x + 84 = 0|:6\\\\x^2 - 9x + 14=0\\\\D = 81 - 56 = 25\\\\\sqrt D = 5\\\\ x_1 = \frac{9-5}{2} = \frac{4}{2} = 2\\\\ x_2 = \frac{9+5}{2} = \frac{14}{2} = 7$

3а)

$\frac{x^2-x}{3} = \frac{7x-8}{9} \\\\ 3x^2 - 3x = 7x - 8\\\\3x^2 - 3x - 7x + 8 = 0\\\\3x^2 - 10x + 8 = 0\\\\D = 100-96= 4\\\\\sqrt D = 2\\\\ x_1 = \frac{10-2}{6} = \frac{8}{6} = 1\frac{2}{6} = 1\frac{1}{3}\\\\x_2 = \frac{10+2}{6} = \frac{12}{6} = 2$

3б)

$\frac{x^2-1}{6} - 6x = 6 |\cdot 6\\\\x^2-1 - 36x - 36 = 0\\\\x^2 - 36x - 37 = 0\\\\D = 1296 + 148 = 1444\\\\ \sqrt D = 38\\\\x_1 = \frac{36-38}{2} = \frac{-2}{2} = -1\\\\x_2=\frac{38+36}{2} = \frac{74}{2} = 37$

3в)

$\frac{x^2+3x}{4} = \frac{x^2+16}{8}\\\\ 2x^2 + 6x = x^2 + 16\\\\2x^2 - x^2 + 6x -16 = 0\\\\x^2 + 6x - 16 = 0\\\\D = 36 + 64 = 100\\\\\sqrt D = 10\\\\x_1 = \frac{-6-10}{2} = \frac{-16}{2} = -8\\\\x_2 = \frac{-6+10}{2} = \frac{4}{2} = 2$

3г)

$\frac{9x^2+x}{2} - \frac{6-4x}{3} = \frac{2x^2+11}{3}\\\\3(9x^2+x) - 2(6-4x) = 2(2x^2 +11) \\\\ 27x^2 + 3x - 12 + 8x - 4x^2 - 22 = 0\\\\ 23x^2 + 11x - 34 = 0\\\\D = 121 + 3128 = 3249\\\\\sqrt D = 57\\\\x_1 = \frac{-11-57}{46} = \frac{-68}{46} = \frac{-34}{23}\\\\x = \frac{-11+57}{46} = \frac{46}{46} = 1$

Удачи!