Задание 19, стр 98

$\begin{array}{ccc}\hfill \text{a})& x^2-2x+3=0.\hfill & \hfill \\ \hfill & D=b^2-4\mathit{ac}=4+12=16=4^2.\hfill \\ \hfill & x_1=\frac{-b+\sqrt{D}}{2a}=\frac{2+4}{2}=3;x_2=\frac{-b-\sqrt{2}}{2a}=\frac{2-4}{2}=\frac{-2}{2}=-1.\hfill \\ \hfill & x^2-2x-3=(x-3)(x+1).\hfill \\ \hfill & x^2+2x+1={(x+1)}^2\hfill \\ \hfill & \frac{x^2-2x+3}{x^2+2x+1}=\frac{(x-3)(x+1)}{{(x+1)}^2}=\frac{x-3}{x+1}.\hfill \\ \hfill \text{ b) }& x^2-4x+4={(x-2)}^2.\hfill \\ \hfill & x^2+2x-8=0\hfill \\ \hfill & D=b^2-4\mathit{ac}=2^2-4\cdot (-8)=4+32=3b=6^2\hfill \\ \hfill & x_1=\frac{-b+\sqrt{2}}{2a}=\frac{-2+6}{2}=\frac{4}{2}=2;x_2=\frac{-2-6}{2}=\frac{-8}{2}=-4.\hfill \\ \hfill & \frac{x^2-4x+4}{x^2+2x-8}=\frac{{(x-2)}^2}{(x-2)(x+4)}=\frac{x-2}{x+4}.\hfill \\ \hfill \text{ c) }& \frac{x^2-6x+9-y^2}{x^2-3x+3x}=\frac{{(x-3)}^2-y^2}{x(x-3)+\mathit{xy}}=\frac{(x-3-y)(x-3+y)}{x(x-3+y)}=x-3-y.\hfill \end{array}$