Предмет: Алгебра
Автор: Swift14
Вопрос задан 7 лет назад

Помогите решить задание 1

Приложения:

Ответы

Ответ дал: drakerton

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)

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

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)

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