Предмет: Алгебра, автор: sorry123123

Рациональные уравнения (8 класс)

Приложения:

Ответы

Автор ответа: elenamuraweva
1

Ответ:

 \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

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