Question

Răspunsul va rog mult..
Dau coroana..

Answer 1

$\frac{2(x-3)+x+2}{(x+2)*(x+3)} = \frac{2x+6+x+2}{(x+2)*(x+3)} = \frac{3x+8}{x+2*(x+3)}$[/tex]a)
tex] \frac{2(x-3)+x+2}{(x+2)*(x+3)} = \frac{2x+6+x+2}{(x+2)*(x+3)} = \frac{3x+8}{x+2*(x+3)} [/tex]
b)
[tex](x+2)* \frac{1}{x+4} = \frac{x+2}{x+4} [/tex]
c)
[tex] \frac{x-3}{x-4}* \frac{x-5}{x-3} = \frac{1}{x-4}* (x-5) = \frac{x-5}{x-4} [/tex]

Answer 2

2/x+2 + 1/x+3 =  2(x+3)+(x+2)/(x+2)(x+3)=2x+6+x+2/(x+2)(x+3)=3x+8/(x+2)(x+3)

x+2/x+4

x-3/x-4  *  x-5/x-3=x-5/x-4