a) E(x)= [tex][ \frac{ x^{2} }{x(x-4)} + \frac{ (x-4)^{2} }{x(x-4)}- \frac{x(x-4)}{x(x-4)}]* \frac{x(x-2)}{ x^{2} -4x+16}= \\ = \frac{ x^{2} + x^{2} -8x+16- x^{2} +4x}{x(x-4)}* \frac{x(x-2)}{ x^{2} -4x+16}= \\ = \frac{ x^{2}-4x+16 }{x(x-4)}* \frac{x(x-2)}{ x^{2} -4x+16} = \\ = \frac{x-2}{x-4} [/tex]
b) Pentru ca E(x) >0 este necesar ca (x-2) si (x-4) sa fie simultan pozitive sau simultan negative.
In primul caz avem x∈(4;+∞), iar in al doilea caz avem x∈(-∞;2).
c)[tex]E(a)= \frac{a-2}{a-4} [/tex]∈Z => a-4 | a-2 ,dar a-4 | a-4 => a-4 | (a-2)-(a-4) <=> a-4 | 2.
(a-4)∈[tex] D_{2} [/tex]={-2,-1,1,2}
a-4=-2 => a=2
a-4=-1 => a=3
a-4=1 => a=5
a-4=2 => a=6
a∈{2,3,5,6}
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Daca intr-adevar era 2x, atunci am fi avut:
[tex]E(x)= \frac{2}{x-4} [/tex]
b) Fractia era mai mare decat 0 daca x-4>0, deci x∈(4;+∞).
c)Am fi avut a-4 |2 => (a-4)∈[tex] D_{2} [/tex]={-2;-1;1;2}.
Se obtinea a∈{2;3;5;6}.
