Answered

At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Explore our Q&A platform to find in-depth answers from a wide range of experts in different fields. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Simplify this expression: x^2+6x+5/x^2-25

Sagot :

[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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.