Variable squared minus negative 2
Answer:
5
Step-by-step explanation:
AC=AB+BC
so 16=x+1+x+7
which simplifies to 16=2x+8
subtract eight from both sides to get 8=2x
then divide by 2 to get that x=4
AB=x+1, which substitutes into 4+1=5
Consider expression 
First, you can factor it:
Since x is integer number, then you can see that x-1 is previous integer number (x-1 is 1 unit smaller than x).
Therefore, x-1 and x are two consecutive integers. When you have two consecutive integers, one of them is always even and one is always odd. Multiplying even integer number by odd integer number you always get even integer number.
Thus, 
is always even.
5^(x+7)=(1/625)^(2x-13)
We move all terms to the left:
5^(x+7)-((1/625)^(2x-13))=0
Domain of the equation: 625)^(2x-13))!=0
x∈R
We add all the numbers together, and all the variables
5^(x+7)-((+1/625)^(2x-13))=0
We multiply all the terms by the denominator
(5^(x+7))*625)^(2x+1-13))-((=0
We add all the numbers together, and all the variables
(5^(x+7))*625)^(2x-12))-((=0
We add all the numbers together, and all the variables
(5^(x+7))*625)^(2x=0
not sure if this is right :/
