Question

Cine ma poate ajuta macar un exercitiu
dau coroana va rog urgent

Answer 1

$5.a) \frac{y+2-(2y-3)}{2y+1}= \frac{y+2-2y+3}{2y+1}= \frac{5-y}{2y+1} \\ \\ \\ b) \frac{4(x+1)+3(2-x)}{12x}= \frac{4x+4+6-3x}{12x}= \frac{x+10}{12x} \\ \\ c)= \frac{(x-4)(x+4)}{2(x+3)}* \frac{(x+3)^{2} }{x+4} = \frac{(x-4)(x+3)}{2} \\ \\ \\ d) \frac{y(y+3)}{y-1}: \frac{(y-3)(y+3)}{(y-1)^{2} } = \\ \\ \frac{y(y+3)}{y-1}* \frac{(y-1) ^{2}}{(y-3)(y+3)}=$

6.a=( x+4)/(x+2)=(x+2+2)/(x+2)=(x+2)/(x+2)+2)/(x+2)=1+2/(x+2)
a=1+2/(x+2)  pentru x=0 se obtine a=1+2/2=1+1=2∈N

pentru x=0⇒a=2∈N

Answer 2

5)
(y+2)/(2y+1)-(2y-3)/(2y+1)=(y+2-2y+3)/(2y+1)=(5-y)/(2y+1)

(x+1)/3x+ (2-x)/4x=(4x+4+6-3x)/12x=(x+10)/12x

(x²-16)/(2x+6)× (x²+6x+9)/(x+4)=(x-4)(x+4)/2(x+3)× (x+3)²/(x+4)=
=(x-4)/2  (x+3)/1=(x-4)(x+3)/2

(y²+3y)/(y-1) : (y²-9)/(y²-2y+1)= y(y+3)/(y-1) : (y-3)(y+3)/(y-1)²=
=y(y+3)/(y-1) × (y-1)²/(y-3)(y+3)= y/1 × (y-1)/(y-3)= y(y-1)/(y-3)

6)
a=( x+4)/(x+2)=[(x+2)+2]/(x+2)=1+ 2/(x+2)
ptr ca 1+ 2/(x+2) sa apartina lui N trebuie ca x+2=2  ⇒ x=0


pentru x=0⇒a=2∈N