Question

Urgent 55 de puncte clasa V


Transformaţi următoarele fracţii zecimale în fracţii ordinare ireductibile:


I.

a) 0,(6) ; b) 0,(27) ; c) 0,(132) ; d) 3,(15) ; e) 12,(243)


II.

a) 0,3(15) ; b) 0,25(7) ; c) 1,34(72) ; d) 23,4(81)

Answer 1

Răspuns:

$I.~a)~0,(6)=\dfrac{6}{9}=\dfrac{2}{3};~~b)~0,(27)=\dfrac{27}{99}=\dfrac{3}{11} ;~~c)~0,(132)=\dfrac{132}{999}=\dfrac{44}{333};~~d)~3,(15)=3\dfrac{15}{99}=3\dfrac{5}{33} =\dfrac{3*33+5}{33}=\dfrac{104}{33} ;~~e)~12,(243)=12\dfrac{243}{999}=12\dfrac{27}{111} =12\dfrac{9}{37}=\dfrac{12*37+9}{37}=\dfrac{453}{37} .$

Explicație pas cu pas:

$II.~0,3(15)=\dfrac{315-3}{990} =\dfrac{312}{990}=\dfrac{52}{165} ;~~b)~0,25(7)=\dfrac{257-25}{900} =\dfrac{232}{900}=\dfrac{58}{225};~~c)~1,34(72)=1\dfrac{3472-34}{9900}=1\dfrac{3438}{9900}=1\dfrac{382}{1100} =1\dfrac{191}{550} =\dfrac{1*550+191}{550}=\dfrac{741}{550};~~d)~23,4(81)=23\dfrac{481-4}{990} =23\dfrac{477}{990}=23\dfrac{53}{110}=\dfrac{23*110+53}{110} =\dfrac{2583}{110} .$

Answer 2

Răspuns:

Explicație pas cu pas:

0,( 6)=9/6=1/3

0,(27)=27/99=9/11

0,( 132)=132/999=44/333

3,(15)=(315 -3)/99=312/99= 104/33

12,( 243) = (12243-12 )/999=4077/333=1359/111

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II

a) 0,3 (15)=(315-3)/990=312/990=156/495=52/165

b) 0,25 (7)= (257-25) /900=232/900=116/450 =58/225

c) 1,34 (72) = (13472 -134 )/9900=13338 /9900=6669/4950=2223/1650=741/550

d) 23,4 (81)=(23481- 234)/990=7739 /330