Hello , help please I need to solve those equations :

1)

2)

3)

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- April 6th 2009, 03:42 PMlinda2005Equations
Hello , help please I need to solve those equations :

1)

2)

3) - April 6th 2009, 05:02 PMMush
- April 6th 2009, 05:14 PMDeMath
- April 6th 2009, 05:26 PMMush
- April 6th 2009, 05:50 PMMush
For the 2nd one, think of it like this:

where k is an integer, and p is a non-integer between 0 and 1.

In mathematical terms:

Hence:

On the RHS, we know that k is the integer part of the number so:

In other words, the NON-integer part of the LHS must be

So now you need to come up with values of k and p such that:

For example, p = 0.1, k = 1 is a solution. (x = 1.1, in other words). - April 6th 2009, 05:58 PMMush
A fairly obvious solution to the last one is x = 0.

- April 7th 2009, 03:18 AMlinda2005
hello thanks Mush , but with x=4 it's ok? so you are wrong

- April 7th 2009, 06:18 AMMush
- April 7th 2009, 10:37 AMNonCommAlg
i'm not sure about 1) but i guess there is no solution. my solutions to 2) and 3):

clearly any integer is a solution. so we'll assume that let then and which is equivalent to

__case 1__. : in this case, since we'll have: but from (1) we have this contradiction means there's no solution in this case.

__case 2.__: then the conditions and (1) are equivalent to conversely if for some positive integer then

and thus and

so the complete set of solutions of the equation is:

Quote:

3)

let where solving the inequalities in (2) for gives you: which gives us thus - April 7th 2009, 01:17 PMlinda2005
Hello, Thanks you NonCommAlg Your solutions is (Yes) , so you can help me with 1) thanks

- April 7th 2009, 01:54 PMNonCommAlg
- April 7th 2009, 02:32 PMMush
There is no solution

Using the following two properties:

It can be deduced that:

Hence:

And in general

Continuing like this, the equation can be reduced to:

It is non-integer, and so there are no solutions. - April 7th 2009, 02:50 PMNonCommAlg
- April 7th 2009, 03:03 PMMush
- April 7th 2009, 03:17 PMlinda2005
look this

When he plugs , the result is .We're going to need to take off three. Now imagine we decrease that slightly. Every term on the lefthand side will decrease by AT LEAST one (for example, ), therefore the sum will decrease by AT LEAST six, and that's too much.

some one can more explain this way