Question

Demonstrati ca     E(x)=$\frac{2X-1}{X-1} : (1+ \frac{3x^{2} }{ x^{2} -1} )$   POATE LUA VALOAREA $\frac{x+1}{2x+1}$.

Answer 1

 [tex] E(x)=\frac{2X-1}{X-1} : (1+ \frac{3x^{2} }{ x^{2} -1} )= ...\\
(1)... 1+\frac{3x^2}{x^2-1}=\frac{x^2-1+3x^2}{(x+1)(x-1)}=
=\frac{4x^2-1}{x^2-1}\surd\surd\;;\\
(2)... E(x) = \frac{2x-1}{x-1}*\frac{(x+1)(x-1)}{4x^2-1}=\\
=\frac{(2x-1)(x+1)}{4x^2-1}=\frac{2x^2-x+2x-1}{4x^2-1}=\\
 =\frac{2x^2+x-1)}{4x^2-1}=\frac{(x+1)(2x-1)}{(2x+1)(2x-1)}= (x+1) / (2x+1)}  [/tex]