Question

Fie expresia:
E(x)=$\frac{8x-16}{ x^{2}+x+1 }: \frac{ x^{2}-4x+4}{ x^{3}-1 }$

a)Aflati DVA al expresiei E(x).
b)Aduceti expresia la forma cea mai simpla.
c)Calculati:E($\frac{1}{2}$
d)Determinati pentru care valori naturale ale lui x valoarea lui E(x) este numar natural.

Answer 1

a) E(x) nu este definit daca$x^{2} +x+1=0$, $x^{2} -4x+4=0$ sau $x^{3}-1=0$.
Prima ecuatie nu are solutii in R.
$x^{2} -4x+4=0 <=> (x-2)^{2}=0 => x-2=0=>x=2$
$x^{3}-1=0 => x^{3}=1 =>x=1$
D=R-{1;2}.

b) $E(x)= \frac{8(x-2)}{ x^{2} +x+1}* \frac{ x^{3} -1}{ x^{2} -4x+4}= \\ = \frac{8(x-2)}{ x^{2} +x+1}* \frac{(x-1)( x^{2} +x+1)}{(x-2)^{2} } = \\ = \frac{8(x-1)}{x-2}$

c) $E( \frac{1}{2})= \frac{8( \frac{1}{2}-1) }{ \frac{1}{2}-2 }= \frac{8*(- \frac{1}{2} )}{- \frac{3}{2} }= \frac{-4}{ -\frac{3}{2} } = \frac{8}{3}$

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)∈$D_{8}$={-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}.