`1/(a)+1/(a+b)+1/(a+b+c)=1 ` Tìm` a,b,c` thuộc `N*`

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1. Giải đáp:

    

   Lời giải và giải thích chi tiết:
   – Nếu $a=1=>{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{a+b}}+{\displaystyle \frac{1}{a+b+c}}$

   $=1+{\displaystyle \frac{1}{a+b}}+{\displaystyle \frac{1}{a+b+c}}>1$ (Ko TM)

   – Nếu $a>3=>a+b>3;a+b+c>3$  

   $=>{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{a+b}}+{\displaystyle \frac{1}{a+b+c}}$

   $<{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}<1$ (Ko TM)

    Vậy chỉ có thể $a=2;a=3$
   – Xét $a=2$
   $=>{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{a+b}}+{\displaystyle \frac{1}{a+b+c}}=1$
   

   $<=>{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2+b}}+{\displaystyle \frac{1}{2+b+c}}=1$

   $<=>{\displaystyle \frac{1}{2+b}}+{\displaystyle \frac{1}{2+b+c}}={\displaystyle \frac{1}{2}}(1)$

   + Nếu $b=0$

   $(1)<=>{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{2+c}}={\displaystyle \frac{1}{2}}$

   $<=>{\displaystyle \frac{1}{2+c}}=0$ (vô lý)
   

   + Nếu $b=1$
   $(1)<=>{\displaystyle \frac{1}{2+1}}+{\displaystyle \frac{1}{2+1+c}}={\displaystyle \frac{1}{2}}$

   $<=>{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3+c}}={\displaystyle \frac{1}{2}}$

   $<=>{\displaystyle \frac{1}{3+c}}={\displaystyle \frac{1}{6}}$

   $<=>3+c=6<=>c=3$
   

   + Nếu $b>=2=>b+c>2$

   $(1)<=>{\displaystyle \frac{1}{2+b}}+{\displaystyle \frac{1}{2+b+c}}$

   $<{\displaystyle \frac{1}{2+2}}+{\displaystyle \frac{1}{2+2}}={\displaystyle \frac{1}{2}}$ (Ko TM)

   – Xét $a=3=>a+b>=3;a+b+c>=3$
   

   $=>{\displaystyle \frac{1}{a}}+{\displaystyle \frac{1}{3+b}}+{\displaystyle \frac{1}{3+b+c}}=1$
   

   $=<{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{3}}$

   $=>a=3;b=c=0$

   KL $(a;b;c)=(2;2;0);(2;1;3);(3;0;0)$

    

    

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