The answer is: Â "Â
"Â Â .
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Given: Â 0.1212121212..... repeating ; Â write that value as a fraction;
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In other words; we are given: Â "0.1212121212..... repeating infinitely" ; Â
→ that is to say; "0.12 ...  ; {the "12" decimal portion repeats infinitely} ;Â
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→ We write this value, as a fraction, as:  "12/99" ;
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→Explanation:
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Note:Â "0.99999999...... repeating infinitely; Â = Â "1" .
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→Since:
Let us say that we have:Â
"x = 0.999999 ; repeating infinitely; Â
In order words, let us say we have: "x = 0.9.... ; Â the "9" decimal repeats infinitely;Â
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  Then "10x" ;  (that is: "10" multlipled by "x";  or "10*x" or "10x" );  is equal to:
"10" * (0.999999.....) Â = 9.99999999...... (the "9" decimal repeats infinitely);
in other words: Â 10x = 9.99999999....
Divide each side by "10" ;
to get "x = 0.9999999....." ; the decimal "9" repeats indefinitely...." ;
But if you have: Â "10x = 10" ; Â divide each side of the equation by "10" ;Â
  you get: "x = 1" .Â
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Also, Â if "x = 0.9999...(repeating infinitely);Â
then: Â 10x = 9.99999.
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      10x  =  9.999999999999999......
    −   x  =  0.999999999999999.......
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      9x  =  9.00000000000000000000.....
 →  9x = 9 ;Â
Divide each side of the equation; to get;Â
 9x /9 = 9/ 9 ;  to get:  x = 1 ; and we have: x = 0.9999.... ;  so
 x= 0.99999.... = 1 ;Â
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So, if the numbers "12" is repeating, we divie "12" by "99" ;Â
 that is; we divide "12" by "two 9's" ;  since "12" is a "TWO-digit number"; a "two-digit number" is being repeated infinitely.
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      So;  0.12121212.....(the "12" is the decimal that repeats infinitely);      Â
         Â
= Â 12/99 ; Â which can be simplified;
Divide each side (both the numerator AND the denominator); by "3" ;
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 " 12/99 "  =  "(12÷3) / (99÷3) = 4/33 " .
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The answer is: Â
  .
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