Question

Se consideră expresia E(x) = (x+2\3-3\x+2):x^2+6x+5\3x+6+2\x+1, unde x€R\{-2;-5;-1}
a)CalculeazăE(0)
b)Arată căE(x)=1,pentru oricex€R\{-5;-2;-1}​

Answer 1

Răspuns:

Explicație pas cu pas:

Răspuns atașat ...

Answer 2

$\it a)\ E(0)= (\dfrac{^{2)}2}{\ 3}-\dfrac{^{3)}3}{\ 2}):\dfrac{5}{6}+2=-\dfrac{5}{6}:\dfrac{5}{6}+2=-1+2=1$

$\it b)\ E(x)=\dfrac{(x+2)^2-3^2}{3(x+2)}\cdot\dfrac{3(x+2)}{x^2+5x+x+5}+\dfrac{2}{x+1}=\\ \\ \\ =\dfrac{(x+2-3)(x+2+3)}{x(x+5)+(x+5)}+\dfrac{2}{x+1}=\dfrac{(x-1)(x+5)}{(x+5)(x+1)}+\dfrac{2}{x+1}=\\ \\ \\ =\dfrac{x-1}{x+1}+\dfrac{2}{x+1}=\dfrac{x-1+2}{x+1}=\dfrac{x+1}{x+1}=1$