Question

Решите неравенство:


x² (5 - x)/ x² - 18x + 81 =<0

Answer 1

Ответ:

$\dfrac{x^2(5-x)}{x^2-18x+81}\leq 0\ \ \ \to \ \ \ \ \dfrac{x^2(x-5)}{(x-9)^2}\geq 0\\\\\\x_1=0\ ,\ x_2=5\ ,\ x_3=9\\\\znaki:\ \ ---[\ 0\ ]---[\ 5\ ] +++(9)+++\\\\x\in \{\, 0\, \}\cup [\ 5\ ;\ 9\ )\cup (\ 9\ ;+\infty \, )$