Question

Va multumesc!!!!!............................

Answer 1

a)Stabilim domeniul de definitie: x+2 !=0 =>x!=-2;
x+3!=0=>x!=-3
(x+4)(x+3) =>x!=-3 si -4 
(x+4)(x-6)!=0 =>x!= -4 si 6
Fractia poate lua orice valoare in afara de cele respinse de domeniul de definitie.
b)$(\frac{2}{x+2}+ \frac{x}{x+3} - \frac{x(x+5)}{x ^{2} +5x+6} ): \frac{x^2-2x-24}{ x^{2} +7x+12}$=>aducem primele  doua fractii la numitorul comun :$( \frac{2(x+3)}{(x+2)(x+3)}+ \frac{x(x+2)}{(x+2)(x+3)}- \frac{x(x+5)}{(x+2)(x+3)} ): \frac{(x+4)(x-6)}{(x+4)(x+3)}=\ \textgreater$=>$\frac{2x+6+x^{2}+2x-x^{2}-5x}{(x+2)(x+3)} * \frac{(x+3)(x+4)}{(x+4)(x-6)}=\ \textgreater \ \frac{-x+6}{(x+2)(x+3)} * \frac{(x+3)(x+4)}{(x+4)(x-6)} =\ \textgreater$ 
Se simplifica -x+6 cu x+6;x+3 cu x+3 ; x+4 cu x+4 si obtinem :
$- \frac{1}{x+2}$
c)
9*[tex] (-\frac{1}{x+2} )= -\frac{9}{x+2}=\ \textgreater \ x+2=9|-2 =\ \textgreater\ x=7 [/tex]   CA 9*E(X);  sa apartina lui Z, X trebuie sa fie  7.