Question

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

Answer 1

   
[tex]\displaystyle E(x)= \left( \frac{x}{x -2 } - \frac{1}{4- x^{2} } + \frac{3x}{x +2 } \right) \cdot \frac{ x^{2} +4x +4}{4 x^{2} -1} \cdot \frac{2x +1}{x + 2} = \\ \\ =\left( \frac{x}{x -2 } + \frac{1}{x^{2} -4} + \frac{3x}{x +2 } \right) \cdot \frac{ (x+2)^2}{4 x^{2} -1} \cdot \frac{2x +1}{x + 2} = \\ \\ =\left( \frac{x(x+2)}{x^2 -4 } + \frac{1}{x^{2} -4} + \frac{3x(x-2)}{x^2 -4 } \right) \cdot \frac{ (x+2)^2}{(2x-1)(2x+1)} \cdot \frac{2x +1}{x + 2} = [/tex]


[tex]\displaystyle \\ =\frac{x(x+2) +1 + 3x(x-2)}{x^2 -4 } \cdot \frac{ x+2}{2x-1} = \\ \\ =\frac{x^2+2x +1 + 3x^2-6x}{(x-2)(x+2)} \cdot \frac{ x+2}{2x-1} = \\ \\ =\frac{4x^2 -4x+1}{x-2} \cdot \frac{1}{2x-1} = \\ \\ =\frac{ (2x)^2 -2\cdor (2x)\cdot 1+1^2}{x-2} \cdot \frac{1}{2x-1} = \\ \\ =\frac{ (2x-1)^2 }{x-2} \cdot \frac{1}{2x-1} = \boxed{ \frac{ 2x-1 }{x-2} }\\ \\ \texttt{cctd}[/tex]