Home/ Intrebari/Q 85099
Urmator
In Process

AniDeȘcoală.ro Latest Intrebari

claudia_cla2011
user (0)
Pe: 2014-01-23T07:22:00+02:00 In: In:

Operatii cu numere rationale pozitive

La unele am stiut dar nu am fost sigura si v-am trecut cat mi-a dat ca sa ma verificati. La altele nu am stiut/nu mi-a dat bine deloc . Uitati-le :

\begin{array}{l} 20.{\rm b) 3,2:}\left[ {{\rm 3,3 - }\left( {{\rm 4,}\left( {\rm 3} \right) - 1,7} \right)] \\ 21.\left( {\frac{1}{3} + 0,25} \right) \cdot \left( {1 + \frac{5}{7}} \right) \\ \frac{{{}^{10)}1}}{3} + \frac{{^{3)} 25}}{{10}} = \frac{{10 + 75}}{{30}} = \frac{{85^{(5} }}{{30}} = \frac{{17}}{6} \\ Deci:\frac{{17}}{6} \cdot \left( {\frac{{^{7)} 1}}{1} + \frac{5}{7}} \right) = \frac{{17}}{6} \cdot \frac{{7 + 5}}{7} = \frac{{17}}{6} \cdot \frac{{12}}{7} = \frac{{17 \cdot 2}}{7} = \frac{{34}}{7}? \\ 25.{\rm }c)0,375 \cdot \left( {0,25 + \frac{1}{6}} \right) \cdot \left( {0,2 + \frac{1}{3}} \right) \\ 29.{\rm Cu ajutorul relatiei: }\frac{{a - b}}{{ab}} = \frac{1}{a} - \frac{1}{b},{\rm calculati sumele :} \\ {\rm A = }\frac{{\rm 2}}{{{\rm 5} \cdot {\rm 7}}} + \frac{2}{{7 \cdot 9}} + ...... + \frac{2}{{23 \cdot 25}}{\rm si B = }\frac{{\rm 4}}{{{\rm 1} \cdot {\rm 5}}} + \frac{4}{{5 \cdot 9}} + ....... + \frac{4}{{81 \cdot 85}} \\ \end{array} \]

Astea au mai ramas din 38 de exercitii. Multumesc.

Similare

5 raspunsuri

  1. nina2012
    user (0)

    la 21. 0,25=25/100 nu 25/10

  2. tata2008
    maestru (V)

    \rm{\bl20.{\rm b) 3,2:}\left[ {{\rm 3,3 - }\left( {{\rm 4,}\left( {\rm 3} \right) - 1,7} \right)]= ...\\ I. transformam fractiile zecimale in fractii ordinare ... \\ 3,2=\frac{32}{10} ; 3,3=\frac{33}{10} ; 4,(3)=\frac{43-4}{9}=\frac{39}{9}=\frac{13}{3} ; 1,7=\frac{17}{10} ; 26,(3)=\frac{263-26}{9}=\frac{237}{9}=79 / 3 ;\\ II. ( ... ) = \frac{13}{3}-\frac{17}{10}=\frac{130-51}{30}=\fbox{79 / 30} ;\\ III. (...) : 79/3 adica 79/30 : 79/3 = \frac{1}{10} ;\\ IV. [ ... ]=33/10-1/10=\fbox{\frac{32}{10}} ;\\ V. 32/10 : 32/10 = \fbox{\fbox{1}} \surd\surd\\ 21. \left( {\frac{1}{3} + 0,25} \right)\cdot \left( {1 + \frac{5}{7}} \right) =(\frac{100+75}{300}) \cdot (\frac{7+5}{7})=\frac{175 \cdot 12}{300 \cdot 7}= \fbox{\fbox{ 1 }} \surd\surd \\ 25.{\rm }c) 0,375 \cdot \left( {0,25 + \frac{1}{6}} \right) \cdot \left( {0,2 + \frac{1}{3}} \right) =\frac{375}{1000}\cdot(\frac{25}{100}+\frac{1}{6})\cdot(\frac{2}{10}+\frac{1}{3})=\frac{375\cdot250\cdot16}{1000\cdot600\cdot30}= \fbox{\fbox{\frac{1}{12}}} \surd\surd ; \\

  4. tata2008
    maestru (V)

    claudia_cla2011 wrote:
    \rm{29. Cu ajutorul relatiei: }\frac{{a - b}}{{ab}} = \frac{1}{a} - \frac{1}{b} \\ calculati\;\; sumele : \\ {\rm A = }\frac{{\rm 2}}{{{\rm 5} \cdot {\rm 7}}} + \frac{2}{{7 \cdot 9}} + ...... + \frac{2}{{23 \cdot 25}}{\rm{ si B = }\frac{{\rm 4}}{{{\rm 1} \cdot {\rm 5}}} + \frac{4}{{5 \cdot 9}} + ....... + \frac{4}{{81 \cdot 85}} \\\]


    \rm{\bl\\A=\frac{1}{5}\underbrace{-\not{\frac{1}{7}} + \not{\frac{1}{7}}-\not{\frac{1}{9}} + \not . . . \not +\not{\frac{1}{23}}}_{=0}-\frac{1}{25}=\frac{1}{5}-\frac{1}{25}= \fbox{\frac{4}{25}} \surd \\ B= \frac{1}{1}-\not{\frac{1}{5}} + \not{\frac{1}{5}}-\not{\frac{1}{9}} + \not . . . \not + \not{\frac{1}{81}}-\frac{1}{85}=\frac{1}{1}-\frac{1}{85}=\fbox{\frac{84}{85}} \surd

  5. claudia_cla2011
    user (0)

    tata2008 wrote: \rm{\bl20.{\rm b) 3,2:}\left[ {{\rm 3,3 - }\left( {{\rm 4,}\left( {\rm 3} \right) - 1,7} \right)]= ...\\ I. transformam fractiile zecimale in fractii ordinare ... \\ 3,2=\frac{32}{10} ; 3,3=\frac{33}{10} ; 4,(3)=\frac{43-4}{9}=\frac{39}{9}=\frac{13}{3} ; 1,7=\frac{17}{10} ; 26,(3)=\frac{263-26}{9}=\frac{237}{9}=79 / 3 ;\\ II. ( ... ) = \frac{13}{3}-\frac{17}{10}=\frac{130-51}{30}=\fbox{79 / 30} ;\\ III. (...) : 79/3 adica 79/30 : 79/3 = \frac{1}{10} ;\\ IV. [ ... ]=33/10-1/10=\fbox{\frac{32}{10}} ;\\ V. 32/10 : 32/10 = \fbox{\fbox{1}} \surd\surd\\ 21. \left( {\frac{1}{3} + 0,25} \right)\cdot \left( {1 + \frac{5}{7}} \right) =(\frac{100+75}{300}) \cdot (\frac{7+5}{7})=\frac{175 \cdot 12}{300 \cdot 7}= \fbox{\fbox{ 1 }} \surd\surd \\ 25.{\rm }c) 0,375 \cdot \left( {0,25 + \frac{1}{6}} \right) \cdot \left( {0,2 + \frac{1}{3}} \right) =\frac{375}{1000}\cdot(\frac{25}{100}+\frac{1}{6})\cdot(\frac{2}{10}+\frac{1}{3})=\frac{375\cdot250\cdot16}{1000\cdot600\cdot30}= \fbox{\fbox{\frac{1}{12}}} \surd\surd ; \\

    La 20. b) Cum ti-a dat 13/3 – 17/10 ?? Nu era 33/10 – 13/3 ?

  6. tata2008
    maestru (V)
    Vezi continutul din paranteza rotunda !
    ” pasul II „. … (4,(3) – 1,7) = …
    Urmareste pasii de rezolvare !
Raspunde

Raspunde