Алгебра: Рациональные уравнения (8 класс)

Рациональные уравнения (8 класс)

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Ответ:

yoruhoshi

13.01.2022 08:34

$\frac{x + 7}{x - 2} - 1 = \frac{36}{ {x}^{2} - 4} \\ \frac{x + 7}{x - 2} - 1 = \frac{36}{(x - 2)(x + 2)} \\ \frac{(x + 7)(x + 2) - (x - 2)(x + 2)}{(x - 2)(x + 2)} = \frac{36}{(x - 2)(x + 2)} \\ {x}^{2} + 2x + 7x + 14 - ( {x}^{2} + 2x - 2x - 4) = 36 \\ {x}^{2} + 9x + 14- {x}^{2} + 4 = 36 \\ 9x + 18 = 36 \\ 9x = 36 - 18\\ 9x = 18 \\ x = 18 \div 9 \\ x = 2$

$\frac{x - 5}{x + 3} + \frac{8}{x - 3} = \frac{43}{ {x}^{2} - 9 } \\ \frac{x - 5}{x + 3} + \frac{8}{x - 3} = \frac{43}{(x - 3)(x + 3)} \\ \frac{(x - 5)(x - 3) + 8(x + 3)}{(x - 3)(x + 3)} = \frac{43}{(x - 3)(x + 3)} \\ {x}^{2} - 3x - 5x + 15 + 8x + 24 = 43 \\ {x}^{2} + 39 = 43 \\ {x}^{2} = 43 - 39 \\ {x}^{2} = 4 \\ x = + - 2$

$\frac{x - 6}{x + 1} + \frac{2 + x}{x - 1} = \frac{8}{ {x}^{2} - 1} \\ \frac{x - 6}{x + 1} + \frac{2 + x}{x - 1} = \frac{8}{(x - 1)(x + 1)} \\ \frac{(x - 6)(x - 1) + (2 + x)(x + 1)}{(x - 1)(x + 1)} = \frac{8}{(x - 1)(x + 1)} \\ {x}^{2} - x - 6x + 6 + 2x + 2 + {x}^{2} + x = 8 \\ 2 {x}^{2} - 4x + 8 = 8 \\ 2 {x}^{2} - 4x = 8 - 8 \\ 2 {x}^{2} - 4x = 0 \\ 2x(x - 2) = 0 \\ 1)2x = 0 \\ x = 0 \\ 2)x - 2 = 0 \\ x = 2$

$\frac{x + 3}{x - 8} - \frac{x + 12}{x + 1} = \frac{99}{(x - 8)(x + 1)} \\ \frac{(x + 3)(x + 1) - (x + 12)(x - 8)}{(x - 8)(x + 1)} = \frac{99}{(x - 8)(x + 1)} \\ {x}^{2} + x + 3x + 3 - ( {x}^{2} - 8x + 12x - 96) = 99 \\ {x}^{2} + 4x + 3 - {x}^{2} + 4x - 96 = 99 \\ 8x - 93 = 99 \\ 8x = 99 + 93 \\ 8x = 192 \\ x = 192 \div 8 \\ x = 24$