от ammornil » 04 Ное 2012, 14:11
[tex]\frac{1}{x-8}-\frac{1}{x-2}=\frac{1}{x-11}-\frac{1}{x-10}[/tex]
ДМ:
[tex]\left|x-8 \ne 0 \\ x-2 \ne 0 \\ x-11 \ne 0 \\ x-10 \ne 0 \right, \hspace{6} \Rightarrow \left| x \ne 8 \\ x \ne 2 \\ x \ne 11 \\ x \ne 10 \right, \hspace{6} \Rightarrow x \in (-\infty;2) \cup (2;8) \cup (8;10) \cup (10;11) \cup (11;+\infty)[/tex]
[tex]\frac{1}{x-8}-\frac{1}{x-2}=\frac{1}{x-11}-\frac{1}{x-10} \\
(x-2).(x-10).(x-11)-(x-8).(x-10).(x-11)=(x-2).(x-8).(x-10)-(x-2).(x-8).(x-11) \\
(x-10).(x-11).[x-2-(x-8)]=(x-2).(x-8).[x-10-(x-11)] \\
(x^2-21.x+110).6=(x^2-10.x+16).1 \\
6.x^2-126.x+660-x^2+10.x-16=0 \\
5.x^2-116.x+644=0 \\
D=58^2-5.644=3364-3220=144 \\
x_{_{1,2}}=\frac{-58 \pm 12}{5} \\
x_{_{1}}=\frac{-70}{5}=-14 \hspace{12} \in \cyr{DM} \\
x_{_{2}}=\frac{-46}{5}=-9\frac{1}{5} \hspace{12} \in \cyr{DM}[/tex]
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]