Нещо съм ги позабравил дробните :(

[Гост]

1. [tex]\frac{2}{11-9x^{2}-30x }[/tex]=[tex]\frac{1}{9x^{2}-6x+1 }[/tex] - [tex]\frac{3}{(3x+11)^2 }[/tex]
2. [tex]\frac{x^{2}-5}{x-1 }[/tex] - [tex]\frac{1-x}{x^{2}-5}[/tex] - [tex]\frac{5}{2 }[/tex] = 0

[Добромир Глухаров]

[tex]2.)\ \frac{x^2-5}{x-1}-\frac{1-x}{x^2-5}-\frac{5}{2}=0[/tex]

[tex]D.C.:\ x\in\mathbb{R}\setminus\{1;\sqrt{5};-\sqrt{5}\}[/tex]

[tex]\frac{x^2-5}{x-1}+\frac{x-1}{x^2-5}=\frac{5}{2}\\\frac{x^2-5}{x-1}=u\\ u+\frac{1}{u}=\frac{5}{2}\\ u^2+1=\frac{5u}{2}\\ 2u^2-5u+2=0\\ (2u-1)(u-2)=0\\ u_1=\frac{1}{2}\Rightarrow \frac{x^2-5}{x-1}=\frac{1}{2}\Rightarrow 2x^2-10=x-1\Rightarrow 2x^2-x-9=0\\ D=1-4.2.(-9)=73\Rightarrow x_{1,2}=\frac{1\pm\sqrt{73}}{4}\\u_2=2\Rightarrow \frac{x^2-5}{x-1}=2\Rightarrow x^2-2x-3=0\Rightarrow x_3=-1;\ x_4=3[/tex]

И четирите получени корена са допустими стойности за даденото уравнение, следователно са решения.

[Гост]

Мерси много!

[Добромир Глухаров]

[tex]1.)\ \frac{2}{11-9x^2-30x}=\frac{1}{9x^2-6x+1}-\frac{3}{(3x+11)^2}\\{-\frac{2}{(3x+11)(3x-1)}=\frac{1}{(3x-1)^2}-\frac{3}{(3x+11)^2}\|.(3x+11)^2.(3x-1)^2}\\\cyr{D.C.}:\ x\in\mathbb{R}\setminus \{-\frac{11}{3};\frac{1}{3}\}\\-2(3x+11)(3x-1)=(3x+11)^2-3(3x-1)^2\\-2(9x^2+30x-11)=9x^2+66x+121-3(9x^2-6x+1)\\\cancel{-18x^2}-60x+22=\cancel{-18x^2}+84x+118\\-144x=96\\x=-\frac{2}{3}\in \cyr{D.C.}[/tex]