Question

помогите срочн000001​

Answer 1

Объяснение:

$y=\sqrt{x^2-14+40}.$

ОДЗ:

$x^2-14x+40\geq 0\\x^2-14x+40=0\\D=(-14)^2-4*1*40=196-160=36\\\sqrt{D}=\sqrt{36} =6.\\x_1=\frac{14-6}{1*2} =\frac{8}{2} =4.\\x_2=\frac{14+6}{1*2} =\frac{20}{2} =10.\ \ \ \ \Rightarrow\\(x-10)*(x-4)=0\\(x-10)*(x-4)\geq 0$

-∞__+__4__-__10__+__+∞

Ответ: x∈(-∞;4]U[10;+∞).

Answer 2

Ответ:

$y=\sqrt{x^2-14x+40}\\\\OOF:\ x^2-14x+40\geq 0\\\\D=36\ \ ,\ \ x_1=4\ ,\ x_2=10\\\\(x-4)(x-10)\geq 0\\\\znaki:\ \ \ +++[\ 4\ ]---[\ 10\ ]+++\\\\x\in (-\infty ;\ 4\ ]\cup [\ 10\ ;+\infty )$