Question

очень . Скоро урок алгебры!

Answer 1

$1)\ \ 9-c^2+a^2-6a=(a^2-6a+9)-c^2=(a-3)^2-c^2=\\\\=(a-3-c)(a-3+c)\\\\2)\ \ x^2-xy+0,25y^2=(x-0,5y)^2\\\\3)\ \ y=3x-1\ \ i\ \ \ \ y=1-x\ \ \ \Rightarrow \ \ \ 3x-1=1-x\ ,\ 4x=2\ ,\ x=0,5\\\\4)\ \ \left\{\begin{array}{l}4x-y=17\ |\cdot 3\\7x+3y=6\end{array}\right\ \oplus \ \left\{\begin{array}{l}y=4x-17\\19x=57\end{array}\right\ \ \left\{\begin{array}{l}y=-5\\x=3\end{array}\right\ \ \ Otvet:\ (3;-5)\ .$

$5)\ \ \left\{\begin{array}{l}x-2y=7\\5x+4y=7\end{array}\right\ \ \left\{\begin{array}{l}x-2y=5x+4y\\x-2y=7\end{array}\right\ \ \left\{\begin{array}{l}-4x=6y\\2y=x-7\end{array}\right\\\\\\\left\{\begin{array}{l}x=-\dfrac{3}{2}\, y\\2y=-\dfrac{3}{2}\, y-7\end{array}\right\ \ \left\{\begin{array}{l}x=-1,5y\\3,5y=-7\end{array}\right\ \ \left\{\begin{array}{l}x=3\\y=-2\end{array}\right\ \ \ \ Otvet:\ (\ 3\ ;-2\ )\ .$