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[tex]E(x)= (\frac{2x}{x+2}+\frac{2x}{6-3x}+\frac{8x}{x^{2}-4 } ):\frac{x-4}{x-2}[/tex]
VA ROGGG


Răspuns :

Rezolvarea este în atașamente.

Vezi imaginea ALEXANDRAVERT
Vezi imaginea ALEXANDRAVERT

[tex]\mathrm{E(x) = \Big(\dfrac{2x}{x+2}+\dfrac{2x}{6-3x}+\dfrac{8x}{x^2-4}\Big):\dfrac{x-4}{x-2}} \\ \\ \mathrm{E(x) = \Bigg(\dfrac{2x}{x+2}+\dfrac{2x}{3(2-x)}+\dfrac{8x}{(x-2)(x+2)}\Bigg)\cdot \dfrac{x-2}{x-4}}\\ \\\mathrm{E(x) = \Bigg(\dfrac{2x}{x+2}-\dfrac{2x}{3(x-2)}+\dfrac{8x}{(x-2)(x+2)}\Bigg)\cdot \dfrac{x-2}{x-4}}[/tex]

[tex]\mathrm{E(x) =\Bigg(\dfrac{2x(x-2)}{x+2}-\dfrac{2x}{3}+\dfrac{8x}{x+2}\Bigg)\cdot \dfrac{1}{x-4}}\\ \\ \mathrm{E(x) = \Bigg(\dfrac{2x^2-4x+8x}{x+2}-\dfrac{2x}{3}\Bigg)\cdot \dfrac{1}{x-4}}\\ \\ \mathrm{E(x) = \Bigg(\dfrac{2x(x+2)}{x+2}-\dfrac{2x}{3}\Bigg)\cdot \dfrac{1}{x-4}} \\ \\ \mathrm{E(x) = \Bigg(2x-\dfrac{2x}{3}\Bigg)\cdot \dfrac{1}{x-4}}\\ \\\\ \mathrm{E(x) = \dfrac{6x-2x}{3(x-4)}} \\ \\ \mathrm{E(x) = \dfrac{4x}{3(x-4)},\quad \forall x\in \mathbb{R}\backslash \{\pm 2, 4\}}[/tex]