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1 answer:
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Answer with Step-by-step explanation:
Let Z be a set of integers
We have to prove that is odd for all value of x
Z.
There are two cases
1.When x is an odd integer number
2.when x is even integer number
1.When x is an odd integer number
When x is positive odd integer
Then is positive odd number (because square of odd positive number is always positive an odd number )
5x is also odd number
-5x -1 = -even number
Suppose x=3
Then -15-1=-16
Odd number - even number=Odd number
Hence , is an odd number .
When x is negative odd number
Then x square is positive odd number and 5x is negative odd term term
Therefore,odd number + odd number -1=Even number -1=Odd number
Hence is an odd number.
2.When x is an even number
When x is positive even number
x square is positive even number and 5x is positive even number
Even number -Even number -1 =Even number -odd number =Odd number
Suppose x=4
=16-20-1=-5=Odd number
When x is negative even number
Then x square is positive even number and 5x is negative even number
=Even number +Even number -1=Even number -1=Odd number
Hence, for all elements of Zis odd.
Answer:
x1 = 2 -
x2= 2 +
Step-by-step explanation:
x/4 + 6 = 7 + 1/4x
(x^2 - 1)/4x = 1
x^2 - 1 = 4x
x^2 - 4x -1 = 0
x= (4+)/2 = 2 -
x= 2 +