Answer :

[tex]\frac{x^2+6x+5}{x^2-25}=(*);\ x^2-25\neq0\to x^2\neq25\to x\neq-5\ \wedge\ x\neq5\\\\x^2+6x+5=0\\\\\Delta=6^2-4\cdot1\cdot5=36-20=16;\ \sqrt\Delta=\sqrt{16}=4\\\\x_1=\frac{-6-4}{2\cdot1}=\frac{-10}{2}=-5;\ x_2=\frac{-6+4}{2\cdot1}=\frac{-2}{2}=-1\\\\x^2+6x+5=(x+5)(x+1)[/tex]


[tex](*)=\frac{(x+5)(x+1)}{(x-5)(x+5)}=\frac{x+1}{x-5}[/tex]
kate200468
[tex] \frac{x^2+6x+5}{x^2-25}= \frac{x^2+6x+5}{(x-5)(x+5)}=(*) \ \ \\ \\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x -5\neq 0\ \ and\ \ x+5 \neq 0\ \ \Rightarrow\ \ x\in R\setminus \{5;-5\}\\ \\x^2+6x+5=(x+5)(x+1)\\ \\because:\\\Delta=6^2-4\cdot1\cdot5=36-20=16\ \ \Rightarrow\ \ \sqrt{\Delta} = \sqrt{16} =4\\ \\ x_1= \frac{-6-4}{2\cdot1} = \frac{-10}{2} =-5,\ \ x_2= \frac{-6+4}{2\cdot1} = \frac{-2}{2} =-1\\ \\ \\ (*)= \frac{(x+5)(x+1)}{(x-5)(x+5)} = \frac{x+1}{x-5} [/tex]