11) αβ>2α−βα+4β multiply by α+4β and β and its >0 because α,β its positive numbers
α(α+4β)>β(2α−β)<=>α2+4αβ>2αβ−β2<=>α2+4αβ−2αβ+β2>0<=>α2+2αβ+β2>0<=>(α+β)2>0
where applicable! So we prove it!
12) Something is wrong here or I don't understand the question:
exp.:32−22=9−4=5 But there is no sum of two consecutive odd numbers be equal with 5 because 1+3 = 4 and general sum of two consecutive odd numbers is allways be equal with even number because if n is a number we have:
n+(n+2)=2n+2 = 2(n+1) for every n natural so we have 2(n+1)
is divided by 2 so it's even number and 5(in this example) it's not even number.
Hope this helps!
Here's 12
Assuming that we are dealing with positive integers....let the odd numbers be
2n - 1 and 2n + 1
Adding these together we get 4n
Sometimes.....in these kinds of proofs, it is good to see if we can discover a pattern....
Let n = 2
And the sum of the odds = 8
And the difference of two squares that equals this is 3^2 - 1^2 = 9 - 1 = 8
We can write this as [ 2n - 1 ]^2 - [ 2n - 3]^2
Note that the numbers in red are just (n - 1) and (n + 1), respectively
Now......Let n = 3
And the sum of the odds = 12
And the difference of two squares that equals this is 4^2 - 2^2 = 16 - 4 = 12
Which we can write as [ 2n - 2 ]^2 - [ 2n - 4 ]^2
Note that, once again, the numbers in red are just ( n - 1) and ( n + 1), respectively
So....it appears that we can write the larger square as [2n - (n- 1)]^2 = [ n + 1]^2
And it appears that we can write the other square as [ 2n - (n + 1) ]^2 = [ n - 1]^2
And their difference is
[ n + 1]^2 - [ n - 1]^2 =
[ n^2 + 2n + 1 ] - [ n^2 - 2n + 1] =
4n