Параметрично уравнение (2x+a^2-a)/(x+2)=2a-x

[Гост]

Дa се реши:
(2x+a^2-a)/(x+2)=2a-x

[ammornil]

[tex]\frac{2.x+a^{2}-a}{x+2}=2.a-x[/tex]

ДМ: [tex]x \ne -2[/tex]

[tex]2.x+a^{2}-a=(2.a-x).(x+2) \hspace{6} \Rightarrow \hspace{6} 2.x+a^{2}-a=2.a.x +4.a -x^{2} -2.x \hspace{6} \Rightarrow \\
2.x+a^{2}-a -2.a.x -4.a +x^{2} +2.x=0 \hspace{6} \Rightarrow \hspace{6} x^{2}-2.(a-1).x+a^{2}-5.a=0[/tex]

[tex]x^{2}-2.(a-1).x+a^{2}-5.a=0 \\
D=(a-1)^{2}-(a^{2}-5.a)=\cancel{a^{2}}-2.a+1\cancel{-a^{2}}+5.a=3.a +1\\
D \ge 0 \hspace{6} \Rightarrow \hspace{6} 3.a+1 \ge 0 \hspace{6} \Rightarrow \hspace{6} a \in [-\frac{1}{3}; +\infty)[/tex]

1случай) ако [tex]a \in (-\infty; -\frac{1}{3}) \hspace{6} \Rightarrow \hspace{6} x \in \oslash[/tex]
2случай) ако [tex]a=-\frac{1}{3} \hspace{6} \Rightarrow \hspace{6} x_{_{1}}=x_{_{2}}=-1\frac{1}{3}[/tex]
3случай) ако [tex]a \in (-\frac{1}{3}; +\infty) \hspace{6} \Rightarrow \hspace{6} x_{_{1}}=a-1 - \sqrt{3.a+1}, \hspace{6} x_{_{2}}=a-1 + \sqrt{3.a+1}[/tex]
[tex]x \ne -2 \hspace{6} \Rightarrow \hspace{6} \left|a-1-\sqrt{3.a+1} \ne -2 \\ a-1+\sqrt{3.a+1} \ne -2 \right, \hspace{6} \Rightarrow \hspace{6} \left|a+1 \ne \sqrt{3.a+1} \\ a+1 \ne -\sqrt{3.a+1} \right, \hspace{6} \Rightarrow \\
\left|a^{2}+2.a+1 \ne 3.a+1 \\ a^{2}+2.a+1 \ne 3.a+1 \right, \hspace{6} \Rightarrow \hspace{6}... \hspace{6} \Rightarrow \hspace{6} \left|a \ne 0 \\ a \ne 1 \right,[/tex]
3.1.) ако [tex]a=0 \hspace{6} \Rightarrow \hspace{6} x_{_{1}}=-2 \notin DM, \hspace{6} x_{_{2}}=2[/tex]
3.2.) ако [tex]a=1 \hspace{6} \Rightarrow \hspace{6} x_{_{1}}=-2 \notin DM, \hspace{6} x_{_{2}}=2[/tex]
3.3.) ако [tex]a \in (-\frac{1}{3};0) \cup (0;1) \cup (1;+\infty) \hspace{6} \Rightarrow \hspace{6} x_{_{1}}=a-1 - \sqrt{3.a+1}, \hspace{6} x_{_{2}}=a-1 + \sqrt{3.a+1}[/tex]

[Гост]

Благодаря!

[deli1]

Има грешка в решението на ammornil

[ammornil]

Да в скобите пред Х трябва да е (а-2)

Благодаря!