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Suzz~Anna
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Pe: 2010-07-11T18:54:34+03:00 In: In:

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Ajutai`ma la Subpunctele: b,c,d,e 😀 va rog Merci>:D<

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7 raspunsuri

  1. s0riyn
    user (0)

    Astea sunt doar niste simple calcule. Cam de cate ori crezi ca ai incalcat regulamentul http://forum.matematic.ro/viewtopic.php?t=3299

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  2. Suzz~Anna
    user (0)

    Am cerut ajutor daca nu vrei sa ma ajuti mai degraba taci

    PS: Nu vreau sa ne certam;)

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  4. Cristian1990
    user (0)

    \rm{\frac{\not5^1}{\not7^1}\cdot\frac{\not{49}^{7}}{\not{75}^{15}}=\frac{7}{15} termenul ...(1); \frac{2}{\not3^1}\cdot\frac{\not9^3}{25}=\frac{6}{25} termenul ...(2); \frac{\not4^2} {\not{11}^1}\cdot\frac{\not{33}^3}{\not{50}^{25}}=\frac{6}{25} termenul ...(3);\\ \frac{7}{15}+\underbrace{\frac{6}{25}-\frac{6}{25}}_{0}=\frac{7}{15}

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  5. Suzz~Anna
    user (0)

    Cristian1990 wrote: \rm{\frac{\not5^1}{\not7^1}\cdot\frac{\not{49}^{7}}{\not{75}^{15}}=\frac{7}{15} termenul ...(1); \frac{2}{\not3^1}\cdot\frac{\not9^3}{25}=\frac{6}{25} termenul ...(2); \frac{\not4^2} {\not{11}^1}\cdot\frac{\not{33}^3}{\not{50}^{25}}=\frac{6}{25} termenul ...(3);\\ \frac{7}{15}+\underbrace{\frac{6}{25}-\frac{6}{25}}_{0}=\frac{7}{15}

    p ala il facusem vroiam doar la b,c,d,e ca la alea nu mio dat:|

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  6. Cristian1990
    user (0)

    \rm{\frac{11}{37}\cdot [(2\frac{1}{2}+1\frac{1}{5})]+(\frac{1}{3}+\frac{5}{6}:\frac{25}{24})\cdot5= ...\\ se efectueaza: (2\frac{1}{2}+1\frac{1}{5})=2+1+\frac{1}{2}+\frac{1}{5}=3+\frac{7}{10}=\frac{37}{10} apoi ...\\ \frac{11}{37}\cdot\frac{37}{10}:\frac{11}{5}=\frac{\not{11}^1}{\not{37}^1}\cdot\frac{\not{37}^1}{\not{10}^2}\cdot\frac{\not5^1}{\not{11}^1}=1/2 *** ;\\ \frac{1}{3}+\frac{\not5^1}{\not6^1}\cdot\frac{\not{24}^4}{\not{25}^2}=\frac{1}{3}+2=7/3 ; \frac{7}{3}\cdot5=35/3 *** ;\\ finalizare ... 1/2+35/3=\frac{3+70}{6}= \frac{73}{3}=\underline{24\frac{1}{3}} .

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  8. Suzz~Anna
    user (0)

    Leam facut pe toate in afara de c:|

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  9. Cristian1990
    user (0)

    Suzz~Anna wrote: Leam facut pe toate in afara de c


    \rm{[...]_1\\ 2\frac{1}{2}-1=1\frac{1}{2}=3/2; \frac{3}{2} \cdot \frac{2}{5} = \frac{3}{5} ;\\ 1,08-2/25=1,08-0,08=1,16=\frac{116^{(4}}{100}=\frac{29}{25};\\ \not2 \cdot \frac{29}{25} \cdot \frac{3}{\not2}=\frac{87}{25} ;\\ [...]_1=\frac{3}{5}+\frac{87}{25}=\frac{15+87}{25}=102/25 *** ;\\ [...]_2\\ 5\frac{1}{3}=16/3 ; 3\frac{3}{4}=15/4 ; \rightarrow \frac{64-45}{12}=19/12 ; \frac{19}{12} \cdot \frac{12}{19}=1 ;\\ [ ... ]_2=1+2=3 *** deci [...]_1 : [...]_2=\frac{102}{25} \cdot \frac{1}{3}=\frac{34}{25}

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