Question

$\frac{x^{2}-9 }{x+1} =\frac{8x}{x+1}$

Answer 1

Ответ:

$\frac{x ^{2} - 9 }{x + 1} = \frac{8x}{x + 1}$

$x + 1 = 0 \\ x + 1 = 0 \\ x + 1 = 0 \\ x + 1 = 0 \\ x = - 1 \\ \frac{x ^{2} - 9 }{x + 1} = \frac{8x}{x + 1} .x≠ - 1$

$\frac{x ^{2} - 9 }{x + 1} = \frac{8x}{x + 1} .x≠ - 1$

${x}^{2} - 9 = 8x$

${x}^{2} - 9 - 8x = 0$

${x}^{2} - 8x - 9 = 0$

${x}^{2} + x - 9x - 9 = 0$

$x(x + 1) - 9(x + 1) = 0$

$(x + 1)(x -9) = 0$

$x + 1 = 0 \\ x - 9 = 0$

$x + 1 = 0 \\ x + 1 - 1 = 0 - 1 \\ x = 0 - 1 \\ x = - 1$

$x - 9 = 0 \\ x - 9 + 9 = 0 + 9 \\ x = 0 + 9 \\ x = 9$

$x = - 1 \\ x = 9.x≠ - 1$

$x = 9$