Question

urgent!!!!! Dau coroană!!!!!!!!​

Answer 1

a) (x+2)(2x-3)=2x²-3x+4x-6=2x²+x-6

b)E(x)=[(x-6)/(x²-25)-x/(5-x)-2/(x+5)]:(2x²+x-6)/(x²-25)=

[(x-6)/(x-5)(x+5)-x/(5-x)-2)(x+5)]:/(x²-25)/(2x²+x-6)=

(x-6+x²+5x-2x+10)/(x-5)(x+5)×(x-5)(x+5)/(x+2)(2x-3)=

(4x²+4x+4)/(x+2)(2x-3)=(x+2)²/(x+2)(2x-3)=(x+2)/(2x-3)

c)E(a)=Z

(a+2)/(2a-3)=±1

a+2=2a-3

a=5

a+2=3-2a

3a=1 nu

(a+2)/(2a-3)=±n N= 1,2,3. ...

a+2=4a-6

3a=8 nu

a+2=6-4a

5a=4 nu