Question

40 баллов Пожалуйста помогите СРОЧНО!!!

Answer 1

Ответ:

$1)\ \ \dfrac{5}{-6x+3}+\dfrac{6x}{1-2x}\geq 0\ \ \ ,\ \ \ ODZ:\ x\ne 0,5\ ,\\\\\\\dfrac{5}{3(1-2x)}+\dfrac{6x}{1-2x}\geq 0\ \ \ ,\ \ \ \dfrac{5+3\cdot 6x}{3(1-2x)}\geq 0\ \ \ ,\ \ \ \dfrac{5+18x}{3(1-2x)}\geq 0\ ,\\\\\\5+18x=0\ \ ,\ \ 18x=-5\ \ ,\ \ x=-\dfrac{5}{18}\\\\1-2x=0\ \ ,\ \ 2x=1\ \ ,\ \ x=\dfrac{1}{2}=0,5\\\\znaki:\ \ \ ---[\, -\frac{5}{18}\ ]+++(\frac{1}{2})---\\\\x\in \Big(-\infty \, ;-\dfrac{5}{18}\ \Big)\cup \Big[\ \dfrac{1}{2}\ ;\, +\infty \, \Big)$

$2)\ \ \dfrac{x^2+7x+8}{(x+1)^2-9}-\dfrac{3x+7}{3x-6}\leq 0\ \ ,\ \ \ ODZ:\ x\ne 2\ ,\ x\ne -4\ ,\\\\\\\star \ \ (x+1)^2-9=(x+1-3)(x+1+3)=(x-2)(x+4)\\\\3x-6=3(x-2)\ \ \star \\\\\\\dfrac{x^2+7x+8}{(x-2)(x+4)}-\dfrac{3x+7}{3(x-2)}\leq 0$

$\dfrac{3(x^2+7x+8)-(x+4)(3x+7)}{3\, (x-2)(x+4)}\leq 0\ \ ,\ \ \dfrac{2x^2+2x-4}{3\, (x-2)(x+4)}\leq 0\\\\\\\dfrac{2\, (x^2+x-2)}{3\, (x-2)(x+4)}\leq 0\ \ ,\ \ \ \dfrac{2\, (x-1)(x+2)}{3\, (x-2)(x+4)}\leq 0\\\\\\znaki:\ \ +++(-4)---[-2\ ]+++[\ 1\ ]---(2)+++\\\\x\in (-4\ ;\, -2\ ]\cup [\ 1\ ;\ 2\ )$