a) E(x) nu este definit daca[tex] x^{2} +x+1=0[/tex], [tex] x^{2} -4x+4=0[/tex] sau [tex] x^{3}-1=0[/tex].
Prima ecuatie nu are solutii in R.
[tex] x^{2} -4x+4=0 <=> (x-2)^{2}=0 => x-2=0=>x=2[/tex]
[tex] x^{3}-1=0 => x^{3}=1 =>x=1 [/tex]
D=R-{1;2}.
b) [tex]E(x)= \frac{8(x-2)}{ x^{2} +x+1}* \frac{ x^{3} -1}{ x^{2} -4x+4}= \\ = \frac{8(x-2)}{ x^{2} +x+1}* \frac{(x-1)( x^{2} +x+1)}{(x-2)^{2} } = \\ = \frac{8(x-1)}{x-2} [/tex]
c) [tex]E( \frac{1}{2})= \frac{8( \frac{1}{2}-1) }{ \frac{1}{2}-2 }= \frac{8*(- \frac{1}{2} )}{- \frac{3}{2} }= \frac{-4}{ -\frac{3}{2} } = \frac{8}{3} [/tex]
d) E(x)∈N=> x-2 | 8(x-1) <=> x-2 | 8x-8.
x-2 | x-2 => x-2 | 8(x-2) <=> x-2 | 8x-16.
x-2 | (8x-8)-(8x-16) <=> x-2 | 8 => (x-2)∈[tex] D_{8} [/tex]={-8;-4;-2;-1;1;2;4;8}.
x∈N => x-2≥-2 => (x-2)∈{-2;-1;1;2;4;8} => x∈{0;1;3;4;6;10}, dar conform punctului a) , x≠1 => Solutia este x∈{0;3;4;6;10}.
