help me please!

[honeypie4321]

i need help with this please

x+7
_______
lxl+lx-1l , and 0<x<1

the whole fraction is in absolute value as well

the answer is x+7 and i dont know why

 

[stapel]

the answer is x+7 and i dont know why

You have posted only an expression, without any instructions. Without knowing what the question is, we can't "know why", either.

Please reply with clarification, including the instructions.

Thank you.

Eliz.

 

[Unco]

i need help with this please

x+7
_______
lxl+lx-1l , and 0<x<1

the whole fraction is in absolute value as well

the answer is x+7 and i dont know why

From 0 < x < 1:

Since x > 0, |x| = ?

Since x < 1, |x - 1| = ?

Then you can simplify \(\displaystyle \mbox{ \left|\frac{x + 7}{|x| + |x - 1|}\right|}\).

 

[honeypie4321]

the instructions ask to rewrite the problem without absolute value symbols

 

[pka]

\(\displaystyle \L
\begin{array}{l}
0 < x < 1\quad \Rightarrow \\
x + 7 = x + 7 \\
|x| = x \\
|x - 1| = 1 - x \\
\quad \Rightarrow \quad \left| {\frac{{x + 7}}{{|x| + |x - 1|}}} \right| = \frac{{x + 7}}{{x + (1 - x)}} = x + 7 \\
\end{array}\)

 

[honeypie4321]

i dont understand where the absolute value signs went.