Question

пожалуйста, очень срочно ​

Answer 1

Ответ:

       

$8)\ \ \displaystyle y=\frac{15-3x}{x^2+2x+1}\ \ ,\ \ \ y=1\\\\\\\frac{15-3x}{x^2+2x+1}>1\ \ ,\ \ \ \frac{15-3x-x^2-2x-1}{(x+1)^2}>0\ \ ,\ \ \frac{-(x^2+5x-14)}{(x+1)^2}>0\ ,\\\\\\\frac{-(x+7)(x-2)}{(x+1)^2}>0\ \ ,\ \ \ \frac{(x+7)(x-2)}{(x+1)^2}<0\ \ ,\\\\\\znaki:\ \ \ +++(-7)---(-1)---(2)+++\\\\\\x\in (-7\ ;\, -1\ )\cup (-1\ ;\ 2\ )$

$9)\ \ \left\{\begin{array}{l}a^2+b^2=25\\a^2+c^2=16\\b^2+c^2=23\end{array}\right\ \oplus \ \left\{\begin{array}{l}2(a^2+b^2+c^2)=64\\a^2+b^2=25\\a^2+c^2=16\\b^2+c^2=23\end{array}\right\ \ \left\{\begin{array}{l}a^2+b^2+c^2=32\\a^2+b^2=25\\a^2+c^2=16\\b^2+c^2=23\end{array}\right$

$(a^2+b^2)+c^2=25+c^2=32\ \ \ \to \ \ \ c^2=7\ \ ,\ \ \ c=\sqrt7\\\\a^2+(b^2+c^2)=a^2+23=32\ \ \ \to \ \ \ a^2=9\ \ \ ,\ \ a=3\\\\(a^2+c^2)+b^2=16+b^2=32\ \ \ \to \ \ \ b^2=16\ \ , \ \ b=4\\\\S_{poln,}=2S_{osn.}+S_{bok.}=2\cdot a\cdot b+2(a+b)\cdot c=2\cdot 3\cdot 4+2(3+4)\cdot \sqrt7=\\\\{}\qquad \qquad \qquad \qquad \qquad =24+14\sqrt7$