Question

ПОМОГИТЕ ПОЖАЛУЙСТА С АЛГЕБРОЙ

Answer 1

$1)\ \ 5x^2+6x+18=0\ \ ,\ \ \ D=b^2-4ac=6^2-4\cdot 5\cdot 18=324<0\\\\\\2)\ \ x^2+3x+2=0\ \ ,\ \ D=3^2-4\cdot 1\cdot 2=9-8=1\ ,\\\\x_1=\dfrac{-b-\sqrt{D}}{2a}=\dfrac{-3-1}{2\cdot 1}=-2\ \ ,\ \ x_2=\dfrac{-b+\sqrt{D}}{2a}=\dfrac{-3+1}{2\cdot 1}=-1\\\\\\3)\ \ x^2+Vx+N=0\ \ ,\ \ \ x_1=-11\ ,\ \ x_2=1\\\\\left\{\begin{array}{l}x_1+x_2=-V\\x_1\cdot x_2=N\end{array}\right\ \ \left\{\begin{array}{l}-11+1=-V\\-11\cdot 1=N\end{array}\right\ \ \left\{\begin{array}{l}V=10\\N=-11\end{array}\right\\\\\\x^2+10x-11=0$

$4)\ \ \dfrac{(x-1)^2}{x^2+2x-3}=\dfrac{(x-1)^2}{(x-1)(x+3)}=\dfrac{x-1}{x+3}\\\\\\5)\ \ x^2-7=12x\ \ \ \to \ \ \ x^2-12x-7=0\\\\D=12^2-4\cdot 1\cdot (-7)=172\\\\x_1=\dfrac{12-\sqrt{172}}{2}=\dfrac{12-2\sqrt{43}}{2}=6-\sqrt{43}\ \ ,\ \ \ x_2=6+\sqrt{43}$