Question

(x^2-15x+50)/(x-4)>0 решить неравенство
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Answer 1

Ответ:

$\dfrac{x^2-15x+50}{x-4}>0\\\\\\\dfrac{(x-5)(x-10)}{x-4}>0\\\\\\znaki\ f(x):\ \ \ ---(4)+++(5)---(10)+++\\\\\\x\in (4,5)\cup (10,+\infty )\\\\\\\star \ \ x^2-15x+50=0\ \ \to \ \ \ x_1=5\ ,\ x_2=10\ \ (teorema\ Vieta)\\\\x^2-15x+50=(x-5)(x-10)\ \ \star$