Что тут? Очень срочно нужен ответ. Created by valyaevaalena. algebra-ru

Что тут? Очень срочно нужен ответ...

WednesAddams

$\displaystyle \frac{7+7x-3x^2}{3x^2+8x+4}=\frac{-3x^2+7x+6}{3x^2+6x+2x+4}=\frac{-3x^2+9x-2x+6}{3x(x+2)+2(x+2)}=\frac{-3x(x-3)-2(x-3)}{(x+2)(3x+2)}=\\ \\ \frac{-(x-3)(3x+2)}{(x+2)(3x+2)}= \frac{-(x-3)}{x+2}=-\frac{x-3}{x+2}=\frac{3-x}{x+2}$

2 votes Thanks 1

elenamuraweva

Ответ:

$\frac{6 + 7x - 3 {x}^{2} }{3 {x}^{2} + 8x + 4} \\$

Найдем корни уравнений:

$- 3 {x}^{2} + 7x + 6 = 0 \\ d = {b}^{2} - 4ac = {7}^{2} - 4 \times ( - 3) \times 6 = 49 + 72 = 121 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 7 - \sqrt{121} }{2 \times ( - 3)} = \frac{ - 7 - 11}{ - 6} = \frac{ - 18}{ - 6} = 3 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 7 + \sqrt{121} }{2 \times ( - 3)} = \frac{ - 7 + 11}{ - 6} = \frac{ 4}{ - 6} = - \frac{2}{3}$

$3 {x}^{2} + 8x + 4 = 0 \\ d = {b}^{2} - 4ac = {8}^{2} - 4 \times 3 \times 4 = 64 - 48 = 16 \\ x1 = \frac{ - b - \sqrt{d} }{2a} = \frac{ - 8 - \sqrt{16} }{2 \times 3} = \frac{ - 8 - 4}{ 6} = \frac{ - 12}{ 6} = - 2 \\ x2 = \frac{ - b + \sqrt{d} }{2a} = \frac{ - 8 + \sqrt{16} }{2 \times 3} = \frac{ - 8 + 4}{ 6} = \frac{ - 4}{6} = - \frac{2}{3}$

$\frac{6 + 7x - 3 {x}^{2} }{3 {x}^{2} + 8x + 4} = \frac{-3(x - 3)(x + \frac{2}{3} )}{3(x + 2)(x + \frac{2}{3}) } = \frac{-(x - 3)}{x + 2} \\$

2 votes Thanks 1

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