Cum se rezolva exercițiul 19 va rog
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Răspunsuri la întrebare
Răspuns:
Explicație pas cu pas:
a) {[1/(1+x)-1]:[1-(1-2x²)/(1+x)]·(4x²+4x+1)/((4x²-1)=
[(1-1-x)/(1+x)]:[(1+x-1+2x²)/(1+x)]·(2x+1)²/(2x+1)(2x-1)=
[-x/(1+x)]:[(2x²+x)/(1+x)]·(2x+1)/(2x-1)=[-x/(1+x)]·(1+x)/(2x²+x)]·(2x+1)/(2x-1)=
[-x(1+x)/(1+x)·x(2x+1)]·(2x+1)/(2x-1)=[-x/x(2x+1)]·(2x+1)/(2x-1)= -1/(2x+1)·(2x+1)/(2x-1)=-(2x+1)/(2x+1)(2x-1)=-1/(2x-1)=1/(1-2x)
E(x)=1/(1-2x)
b) E(1)=1/(1-2)=1/(-1)=-1 ⇒1/E(1)=1/(-1)=-1
E(2)=1/(1-4)=1/(-3)=-1/3 ⇒1/E(2)=1/(-1/3)=-3
E(3)=1/(1-6)=1/(-5)=-1/5 ⇒1/E(3)=1/(-1/5)=-5
E(4)=1/(1-8)=1/(-7)=-1/7 ⇒1/E(4)=1/(-1/7)=-7
E(5)=1/(1-10)=1/(-9)=-1/9 ⇒1/E(5)=1/(-1/9)=-9
E(6)=1/(1-12)=1/(-11)=-1 /11 ⇒1/E(6)=1/(-11)=-11
E(7)=1/(1-14)=1/(-13)=-1/13 ⇒1/E(7)=1/(-13)=-13
E(8)=1/(1-16)=1/(-15)=-1/15 ⇒1/E(8)=1/(-15)=-15
E(9)=1/(1-18)=1/(-17)=-1/17 ⇒1/E(9)=1/(-1/17)=-17
E(10)=1/(1-20)=1/(-19)=-1/19 ⇒1/E(10)=1/(-1)/19=-19
S= 1/E(1)+1/E(2)+1/E(3)+1/E(4)+1/E(5)+1/E(6)+....+1/E(10)=-1+(-3)+(-5)+...+(-19)=
(-1)·(1+3+5+...+19)=(-1)·[(1+2+3+...+19)-2-4-6-...-18]=
(-1)·[(19·20)/2-2(1+2+3+...+9)]=(-1)[(19·10)-2(9·10)/2]=(-1)·(190-90)=(-1)·100=-100
S=-100