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Ajutor ! Ofer 99 puncte .... Sa se afle numerele a ,b,c stiind ca a/b+1 =b/ c+1 = c/2 si b este media aritmetica dintre a si c

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Albastruverde12
[tex]Avem~b= \frac{a+c}{2}. \\ \\ \frac{a}{b+1}= \frac{b}{c+1}= \frac{c}{2} \Leftrightarrow \frac{a}{ \frac{a+c}{2}+1 } = \frac{ \frac{a+c}{2} }{c+1}= \frac{c}{2} \Leftrightarrow \frac{a}{ \frac{a+c+2}{2} }= \frac{a+c}{2(c+1)}= \frac{c}{2} \Leftrightarrow \\ \\ \Leftrightarrow \frac{2a}{a+c+2}= \frac{a+c}{2(c+1)}= \frac{c}{2} . \\ \\ Din~\frac{a+c}{2(c+1)} = \frac{c}{2} ~rezulta~a+c= \frac{2c(c+1)}{2} \Leftrightarrow a+c=c^2+c \Rightarrow a=c^2. [/tex]

[tex]Daca~c=0 \Rightarrow a=b=0. \\ \\ Daca~c \neq 0,~atunci,~din~ \frac{2a}{a+c+2}= \frac{c}{2}~rezulta~a+c= \frac{4a}{c}-2 \Leftrightarrow \\ \\ c^2+c= \frac{4c^2}{c}-2 \Leftrightarrow c^2+c =4c-2 \Leftrightarrow c^2=3c-2. \\ \\ Deci~c^2-3c+2=0 \Leftrightarrow (c-1)(c-2)=0 \Rightarrow c \in \{1;2 \}.[/tex]

[tex]c=1 \Rightarrow a=1 ~si~b= \frac{1+1}{2}=1. \\ \\ c=2 \Rightarrow a= 2^2=4 ~si~b= \frac{ 4+2}{2}=3. \\ \\ Solutie:~(a,b,c) \in \{(0,0,0);(1,1,1);(4,3,2)\}.[/tex]
Incognito
[tex]a,b,c\in\mathbb{N}\\ \frac{a}{b+1}=\frac{b}{c+1}=\frac{c}{2}\\ b=\frac{a+c}{2}\Rightarrow a+c=2b\\ Putem\ folosi\ siruri\ de \ rapoarte \egale.\ Adunam\ prima\ cu\ a\ treia\: \\ \frac{a+c}{b+3}=\frac{2b}{b+3}=\frac{c}{2}\Rightarrow\frac{b}{b+3}=\frac{c}{4}\Rightarrow\\ \frac{b+3}{b}=\frac{4}{c}\Rightarrow 1+\frac{3}{b}=\frac{4}{c}\\ Deoarece\ 1+\frac{3}{b}\ \textgreater \ 1\Rightarrow \frac{4}{c}\ \textgreater \ 1\Rightarrow c\ \textless \ 4\\ Laum\ pe\ cazuri: \\ I)\ c=3\Rightarrow b=9,a=15\ solutia\ nu\ convine\\ II) c=2\Rightarrow b=3,a=4 [/tex]
[tex] solutia\ convine.\\ III) c=1\Rightarrow b=1,c=1\ solutia\ convine\\ IV) c=0\Rightarrow b=0,c=0\\ Asadar\ avem\ TREI SOLUTII.\\[/tex]
OBS;cazul IV este incompatibil cu rationamentul realizat (am intors fractiile). Din acest motiv, ar fi trebuit considerat acest caz inainte de a continua rationamentul-cel in care am inversat fractiile.

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