Question

помогите решить правильно уровнение​

Answer 1

$\frac{7}{x-3} + 1 = \frac{18}{x^2 - 6x + 9}$

$\frac{7}{x-3} + 1 = \frac{18}{(x - 3)^2}$

$\frac{1}{x-3} = t$

$\frac{1}{(x-3)^2} = t^2$

$7t + 1 = 18\cdot t^2$

$18t^2 - 7t - 1 = 0$

$D = 7^2 - 4\cdot(-1)\cdot 18 = 49 + 40 + 32 = 89 + 32 = 121 = 11^2$

$t = \frac{7\pm 11}{2\cdot 18}$

$t_1 = \frac{7 - 11}{2\cdot 18} = \frac{-4}{2\cdot 18} = -\frac{1}{9}$

$t_2 = \frac{7+11}{2\cdot 18} = \frac{18}{2\cdot 18} = \frac{1}{2}$

1) $\frac{1}{x-3} = -\frac{1}{9}$

$x-3 = -9$

$x_1 = 3 - 9 = -6$

2) $\frac{1}{x-3} = \frac{1}{2}$

$x-3 = 2$

$x_2 = 3+2 = 5$

Ответ. {-6; 5}.

Answer 2

Объяснение:

$\frac{7}{x-3} +1=\frac{18}{x^2-6x+9} \\\frac{7}{x-3} +1=\frac{18}{(x-3)^2} .$

ОДЗ: х-3≠0     х≠3.

$7*(x-3)+(x-3)^2=18\\7x-21+x^2-6x+9-18=0\\x^2+x-30=0\\D=121\ \ \ \ \sqrt{D}=11.\\x_1=5\ \ \ \ x_2=-6.$

Ответ: x₁=5, x₂=-6.