Răspuns:
a) [tex]E(x)=(\frac{x+1)2x}{x-1}-\frac{x-1)x}{x+1} ):\frac{2x^{2} +6x}{x^{2} +2x+1} =\frac{2x^{2}+2x-x^{2} +x}{(x-1)(x+1)} :\frac{2x(x+3)}{(x+1)^{2} } =[/tex]
[tex]=\frac{x^{2}+3x}{(x-1)(x+1)}*\frac{(x+1)^{2}}{2x(x+3)}=\frac{x(x+3)}{x-1}*\frac{x+1}{2x(x+3)} = \frac{x+1}{2(x-1)}[/tex]
b) [tex]E(x)*E(-x)=\frac{x+1}{2(x-1)} *\frac{-x+1}{2(-x-1)} =\frac{x+1}{2(x-1)} *\frac{-(x-1)}{-2(x+1)}= \frac{1}{4}[/tex] nu depinde de x
Explicație pas cu pas:
