Question

Rezolvarea va rog....

Answer 1

Răspuns:

Vezi atasanent

Answer 2

$\mathrm{E(x) = \Bigg(\dfrac{x+1}{x-3}-\dfrac{x^2+3x+2}{x^2+4x+3}-\dfrac{1}{9-x^2}\Bigg):\dfrac{x+2}{x^2-9}} \\ \\ \mathrm{E(x) = \Bigg(\dfrac{x+1}{x-3}-\dfrac{(x+1)(x+2)}{(x+1)(x+3)}-\dfrac{1}{(3-x)(3+x)}\Bigg):\dfrac{x+2}{(x-3)(x+3)}}\\ \\ \mathrm{E(x) = \Bigg(\dfrac{x+1}{x-3}-\dfrac{(x+1)(x+2)}{(x+1)(x+3)}+\dfrac{1}{(x-3)(3+x)}\Bigg)\cdot \dfrac{(x-3)(x+3)}{x+2}}\\ \\\\ \mathrm{E(x) =\dfrac{(x+1)(x+3)}{x+2}-(x-3)+\dfrac{1}{x+2}}$

$\mathrm{E(x) = \dfrac{(x+1)(x+3)+1}{x+2}-(x-3)}\\ \\ \mathrm{E(x) = \dfrac{x^2+4x+4}{x+2}-x-3} \\ \\ \mathrm{E(x) = \dfrac{(x+2)^2}{x+2}-x+3} \\ \\ \mathrm{E(x) = x+2-x+3}\\ \\ \mathrm{E(x) = 5}\\ \\ \\ \mathrm{E(m) = 2m+1 \Rightarrow 5 = 2m+1 \Rightarrow 2m = 4 \Rightarrow \boxed{ \mathrm{ m = 2}}}$