Question

помогите решить, кроме №4 и №6​

Answer 1

Ответ:

$1)\ \ \dfrac{a^2+9}{a^2-9}-\dfrac{a}{a+3}=\dfrac{a^2+9-a(a-3)}{(a-3)(a+3)}=\dfrac{12}{(a-3)(a+3)}=\dfrac{12}{a^2-9}\\\\\\2)\ \ 7x^2+9x+2=0\ \ ,\ \ D=25\ \ ,\\\\x_1=\dfrac{-9-5}{14}=-1\ ,\ \ x_2=\dfrac{-9+5}{14}=-\dfrac{2}{7}$

$3)\ \ \left\{\begin{array}{l}2x+3y=-7\\x-y=4\ |\cdot 3\end{array}\right\ \oplus \ \left\{\begin{array}{l}5x=5\\y=x-4\end{array}\right\ \ \left\{\begin{array}{l}x=1\\y=-3\end{array}\right\ \ Otvet:\ (1;-3)\ .$

$5)\ \ y(2)=3\\\\y>0\ \ ,\ esli\ \ x\in (-2\ ;\ 5\ )\\\\y\ vozrastaet\ \ ,\ esli\ \ x\in (-\infty \, ;\ 2\ ]$

$7)\ \ 26\ \ \vee \ \ \sqrt{686}\\\\26^2=676\ <\ 686\ \ \ \Rightarrow \ \ \ \sqrt{676}<\sqrt{686}\ \ \ \Rightarrow \ \ \ \ 26<\sqrt{686}$