Question

Решите пожалуйста 1, 2, 3. Благодарю​

Answer 1

Ответ:

$1)\ \ 5x+1>0\ \ \to \ \ 5x>-1\ ,\ \ x>-0,2\ \ ,\\\\-x+7>0\ \ \to \ \ \ x<7\ \ ,\ \ \underline {-3\in (-\infty ;\, 7\, )\ }\\\\-2x-1<0\ \ \to \ \ \ x>-0,5\\\\x+8<0\ \ \to \ \ \ x<-8$

Ответ: Б) .

$2)\ \ 3x+6<0\ \ \to \ \ \ 3x<-6\ ,\ \ x<-2\ \ ,\ \ \unfrtline {x\in (-\infty ;-2\ )\ }$

Ответ: Б) .

$3)\ \ \left\{\begin{array}{l}x^2+y^2=10\\y-2x=5\end{array}\right\\\\\\(3;1):\ \ \left\{\begin{array}{l}3^2+1^2=10\\1-2\cdot 3=5\end{array}\right\ \ \left\{\begin{array}{l}10=10\\-5\ne 5\end{array}\right$

$(3;-1):\ \ \left\{\begin{array}{l}3^2+(-1)^2=10\\-1-2\cdot 3=5\end{array}\right\ \ \left\{\begin{array}{l}10=10\\-7\ne 5\end{array}\right\\\\\\(-3;1):\ \ \left\{\begin{array}{l}(-3)^2+1^2=10\\1-2\cdot (-3)=5\end{array}\right\ \ \left\{\begin{array}{l}10=10\\7\ne 5\end{array}\right$

$(-3;-1):\ \ \left\{\begin{array}{l}(-3)^2+(-1)^2=10\\-1-2\cdot (-3)=5\end{array}\right\ \ \left\{\begin{array}{l}10=10\\5=5\end{array}\right$

Ответ:  Г) .