Question

Exercitiul 4 din imagine

Answer 1

   
[tex]\displaystyle \\ E(x) = \left( \frac{x-1}{x-2}- \frac{3}{x^2-x-2}- \frac{4}{x+1} \right): \frac{3x}{x+1} = \\ \\ = \left( \frac{x-1}{x-2}- \frac{3}{(x-2)(x+1)}- \frac{4}{x+1} \right) \cdot \frac{x+1}{3x} = \\ \\ = \left( \frac{(x-1)(x+1)}{(x-2)(x+1)}- \frac{3}{(x-2)(x+1)}- \frac{4(x-2)}{(x-2)(x+1)} \right) \cdot \frac{x+1}{3x} = \\ \\ =\frac{(x-1)(x+1)-3-4(x-2)}{(x-2)(x+1)} \cdot \frac{x+1}{3x} = \\ \\ =\frac{ x^2-1 -3-4x+8)}{(x-2)(x+1)} \cdot \frac{x+1}{3x} = [/tex]


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