Question

Решите неравенства методом интервалов:
а)
b) ​

Answer 1

Ответ:

$a)\ \ (3x+9)(x-10)<0\ \ ,\ \ \ x_1=-3\ ,\ x_2=10\\\\znaki:\ \ +++(-3)---(10)+++\\\\x\in (-3\ ;\, 10\ )\\\\\\b)\ \ \dfrac{2x^2-13x+11}{9-x}\geq 0\ \ \ \to \ \ \ \ \dfrac{2x^2-13x+11}{x-9}\leq 0\\\\\\\star \ \ 2x^2-13x+11=0\ \ ,\ \ D=81\ \ ,\ \ x_1=1\ ,\ x_2=\dfrac{11}{2}=5,5\ \ \star \\\\\\\dfrac{2(x-1)(x-5,5)}{x-9}\leq 0\ \ ,\ \ \ ---[\ 1\ ]+++[\, 5,5\, ]---(9)+++\\\\\\x\in (-\infty ;-1\ ]\cup [\ 5,5\ ;\ 9\ )$