Уравнения.

[Гост]

1) [tex]\frac{3x - 1}{x + 7} - \frac{3x^2}{49 - x^2} = \frac{8}{x + 7} + \frac{x - 1}{x - 7}[/tex]

2) [tex]\frac{1 - x^2}{4x^2 - 1} + \frac{5x - 4}{2x + 1} = \frac{8 - x}{1 - 2x}[/tex]

Благодаря предварително! ( :

[ammornil]

[tex]\frac{3.x - 1}{x + 7} - \frac{3.x^2}{49 - x^2} = \frac{8}{x + 7} + \frac{x - 1}{x - 7} \\
\cyr{DM}:
\left|x+7 \ne 0 \\ 49-x^2 \ne 0 \\ x-7 \ne 0 \right, \hspace{12} \Rightarrow \hspace{12} x \in (-\infty;-7) \cup (-7;7) \cup (7;+\infty) \\
\frac{3.x - 1}{x + 7} - \frac{3.x^2}{-(x^2-49)} = \frac{8}{x + 7} + \frac{x - 1}{x - 7} \\
\frac{3.x - 1}{x + 7} + \frac{3.x^2}{(x-7).(x+7)} = \frac{8}{x + 7} + \frac{x - 1}{x - 7} \\
(3.x-1).(x-7)+3.x^2-8.(x-7)-(x-1).(x+7)=0 \\
3.x^2-x-21.x+7+3.x^2-8.x+56-x^2-7.x+x+7=0 \\
5.x^2-36.x+70=0 \\
D=18^2-5.70=324-350<0 \\
\Rightarrow x \in \oslash[/tex]

[ammornil]

[tex]\frac{1 - x^2}{4.x^2 - 1} + \frac{5.x - 4}{2.x + 1} = \frac{8 - x}{1 - 2.x} \\
\cyr{DM}:
\left|4.x^2-1 \ne 0 \\ 2.x+1 \ne 0 \\ 1-2.x \ne 0 \right, \hspace{12} \Rightarrow \hspace{12} x \in (-\infty; -\frac{1}{2}) \cup (-\frac{1}{2};\frac{1}{2}) \cup (\frac{1}{2}; +\infty) \\
\frac{1 - x^2}{(2.x)^2 - 1} + \frac{5.x - 4}{2.x + 1} - \frac{-(x-8)}{-(2.x-1)}=0 \\
\frac{1 - x^2}{(2.x-1).(2.x+1)} + \frac{5.x - 4}{2.x + 1} - \frac{\cancel{-1}.(x-8)}{\cancel{-1.}(2.x-1)}=0 \\
1-x^2+(5.x-4).(2.x-1)-(x-8).(2.x+1)=0 \\
1-x^2+10.x^2-5.x-8.x+4-2.x^2-x+16.x+8=0 \\
7.x^2+2.x+13=0 \\
D=1^2-7.13<0 \\
x \in \oslash[/tex]