Hipi Zhdripi i Matematikës/1064

(b) Nëse a+b<0, atëherë në bazë të relacioneve (6) dhe (a1) del: $\scriptstyle \mid$a+b$\scriptstyle \mid$$\scriptstyle {=}$-(a+b)$\scriptstyle {=}$(-a)+(-b)$\scriptstyle \mid$-a$\scriptstyle \mid$+$\scriptstyle \mid$-b$\scriptstyle \mid$$\scriptstyle {=}$$\scriptstyle \mid$$\scriptstyle \mid$a$\scriptstyle \mid$+$\scriptstyle \mid$b$\scriptstyle \mid$. çka do të thotë se $\scriptstyle \mid$a+b$\scriptstyle \mid$$\scriptstyle \mid$a$\scriptstyle \mid$+$\scriptstyle \mid$b$\scriptstyle \mid$.        Pra, në të dy rastet u vërtetua saktësia e formulës (9).        Kjo teoremë mund të zgjerohet edhe në shumën e n numrave realë, respektivisht: $\scriptstyle {\forall }$a1,a2,...,an$\scriptstyle \in$$\scriptstyle \mathbb {R}$ (9a)        T e o r e m a  4.1.2. -  Vlera absolute e shumës së dy numrave realë a, b nuk është më e vogël se ndryshimi i vlerave absolute të tyre, pra: ($\scriptstyle {\forall }$a,b$\scriptstyle \in$$\scriptstyle \mathbb {R}$)$\scriptstyle \mid$a+b$\scriptstyle \mid$$\scriptstyle \geqslant$$\scriptstyle \mid$a$\scriptstyle \mid$-$\scriptstyle \mid$b$\scriptstyle \mid$. (10)        V ë r t e t i m: Meqë vlen barazia a$\scriptstyle {=}$a+b-b$\scriptstyle {=}$(a+b)+(-b), tani, në bazë të teoremës së mëparshme, përftohet: $\scriptstyle \mid$a$\scriptstyle \mid$$\scriptstyle {=}$$\scriptstyle \mid$(a+b)+(-b)$\scriptstyle \mid$$\scriptstyle \mid$(a+b)$\scriptstyle \mid$+$\scriptstyle \mid$(-b)$\scriptstyle \mid$$\scriptstyle {=}$$\scriptstyle \mid$a+b$\scriptstyle \mid$+$\scriptstyle \mid$b$\scriptstyle \mid$ d.m.th. $\scriptstyle \mid$a$\scriptstyle \mid$$\scriptstyle \mid$a+b$\scriptstyle \mid$+$\scriptstyle \mid$b$\scriptstyle \mid$$\scriptstyle {\Rightarrow }$$\scriptstyle \mid$a$\scriptstyle \mid$-$\scriptstyle \mid$b$\scriptstyle \mid$$\scriptstyle \mid$a+b$\scriptstyle \mid$ .        Nga këto dy teorema del jobarazia e dyfishtë: $\scriptstyle \mid$a$\scriptstyle \mid$-$\scriptstyle \mid$b$\scriptstyle \mid$$\scriptstyle \mid$a+b$\scriptstyle \mid$<$\scriptstyle \mid$a$\scriptstyle \mid$+$\scriptstyle \mid$b$\scriptstyle \mid$. (11)        T e o r e m a  4.1.3. -  Vlera absolute e ndryshimit të çdo dy numrave realë a, b nuk është më e vogël se ndryshimi i vlerave absolute të tyre, pra: ($\scriptstyle {\forall }$a,b$\scriptstyle \in$$\scriptstyle \mathbb {R}$)$\scriptstyle \mid$a-b$\scriptstyle \mid$$\scriptstyle \geqslant$$\scriptstyle \mid$a$\scriptstyle \mid$-$\scriptstyle \mid$b$\scriptstyle \mid$. (12)        V ë r t e t i m: Le të marrim se a-b$\scriptstyle {=}$c, respektivisht se a$\scriptstyle {=}$b+c, atëherë në bazë të teoremës 4.1.1. del: $\scriptstyle \mid$a$\scriptstyle \mid$$\scriptstyle {=}$$\scriptstyle \mid$b+c$\scriptstyle \mid$$\scriptstyle \mid$b$\scriptstyle \mid$+$\scriptstyle \mid$c$\scriptstyle \mid$,ose $\scriptstyle \mid$c$\scriptstyle \mid$$\scriptstyle \geqslant$$\scriptstyle \mid$a$\scriptstyle \mid$-$\scriptstyle \mid$b$\scriptstyle \mid$$\scriptstyle {\Rightarrow }$$\scriptstyle \mid$a-b$\scriptstyle \mid$$\scriptstyle \geqslant$$\scriptstyle \mid$a$\scriptstyle \mid$-$\scriptstyle \mid$b$\scriptstyle \mid$ .        S h e m b u l l i  2 - . Të reduktohet formula f(a)$\scriptstyle {=}$5$\scriptstyle \mid$a$\scriptstyle \mid$+4$\scriptstyle \mid$1-a$\scriptstyle \mid$-7$\scriptstyle \mid$a-2$\scriptstyle \mid$ , ku 2<a<3.        Z g j i d h j e : Meqë nga kushti 2< a< 3 rrjedh 1-a<0 dhe a-2>0, respektivisht $\scriptstyle \mid$1-a$\scriptstyle \mid$$\scriptstyle {=}$-(1-a) dhe $\scriptstyle \mid$a-2$\scriptstyle \mid$$\scriptstyle {=}$a-2, andaj kur këto i zëvendësojmë në formulën e dhënë e marrim: f (a)$\scriptstyle {=}$5a+4 [-(1-a)]-7 (a-2)$\scriptstyle {=}$2 (a+5).

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