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Lusesita
Pe: 2011-09-22T18:48:16+03:00 In: In:

Algebra ! Am nevoie de ajutor urgent!

Nu m-am impacat niciodata cu algebra …sunt la filo , si chiar nu inteleg de ce ne da noua exercitii dinastea!!! Am nevoie de ajutor pana maine!:((((

1) Sa se arate ca urmatoarea expresie este patrat perfect : E =(x+1) totul la patrat , pe langa (x la patrat+1) +x la patrat !

2) se considera nr : x= 29/10 la a treia , y=42/10 la a doua . Sa se scrie sub forma de fractie zecimala nr x+y ; y-x .

3) Calculati :a) |1-2 la puterea n |+ |2 la puterea n -3| , n apartine N
b) |-n-n(n-1)-…-1 / -n-1 | , n apartine N*

Multumesc anticipat :* !

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4 raspunsuri

  1. GreatMath
    veteran (III)

    Lusesita wrote: 1) Sa se arate ca urmatoarea expresie este patrat perfect : E =(x+1) totul la patrat , pe langa (x la patrat+1) +x la patrat !


    Problema 1:

        \[\begin{array}{l} {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) + {x^2} = {x^2}{\left( {x + 1} \right)^2} + {\left( {x + 1} \right)^2} + {x^2} = {\left[ {x\left( {x + 1} \right)} \right]^2} + {\left( {x + 1} \right)^2} + {x^2}\\ = {\left( {{x^2} + x} \right)^2} + 2\left( {{x^2} + x} \right) + 1\\ {\rm{Notam}}:t = {x^2} + x \Rightarrow {\left( {{x^2} + x} \right)^2} + 2\left( {{x^2} + x} \right) + 1 = {t^2} + 2t + 1 = {\left( {t + 1} \right)^2}\\ {\rm{Revin la}}:t = {x^2} + x \Rightarrow {\left( {{x^2} + x} \right)^2} + 2\left( {{x^2} + x} \right) + 1 = {\left( {t + 1} \right)^2} = {\left( {{x^2} + x + 1} \right)^2} \end{array}\]

  2. GreatMath
    veteran (III)

    Problema 3 sub-punctul a

        \[\begin{array}{l} \left| {{\rm{1}} - {{\rm{2}}^{{\rm{n }}}}} \right| = \left\{ \begin{array}{l} 1 - {{\rm{2}}^{\rm{n}}},\,\,n = 0\\ {{\rm{2}}^{\rm{n}}} - 1,\,\,n > 0 \end{array} \right.\\ \left| {{{\rm{2}}^{\rm{n}}} - {\rm{3}}} \right| = \left\{ \begin{array}{l} 3 - {{\rm{2}}^{\rm{n}}},\,\,n \le 1\\ {{\rm{2}}^{\rm{n}}} - 3,\,\,n > 1 \end{array} \right.\\ n = 0 \Rightarrow \left| {{\rm{1}} - {{\rm{2}}^{{\rm{n }}}}} \right| + {\rm{ }}\left| {{{\rm{2}}^{\rm{n}}} - {\rm{3}}} \right| = 1 - {{\rm{2}}^{\rm{n}}} + 3 - {{\rm{2}}^{\rm{n}}} = 2\\ n = 1 \Rightarrow \left| {{\rm{1}} - {{\rm{2}}^{{\rm{n }}}}} \right| + {\rm{ }}\left| {{{\rm{2}}^{\rm{n}}} - {\rm{3}}} \right| = {{\rm{2}}^{\rm{n}}} - 1 + 3 - {{\rm{2}}^{\rm{n}}} = 2\\ n > 1 \Rightarrow \left| {{\rm{1}} - {{\rm{2}}^{{\rm{n }}}}} \right| + {\rm{ }}\left| {{{\rm{2}}^{\rm{n}}} - {\rm{3}}} \right| = {{\rm{2}}^{\rm{n}}} - 1 + {{\rm{2}}^{\rm{n}}} - 3 = 2 \cdot {2^n} - 4 = 4\left( {{2^{n - 1}} - 1} \right) > 0. \end{array}\]

  4. GreatMath
    veteran (III)

    Am sărit peste problema 2 fiindcă chiar nu-mi dau seama care ar fi dificultatea problemei….

  5. edy8
    maestru (V)

    2) se considera nr : x= 29/10 la a treia , y=42/10 la a doua . Sa se scrie sub forma de fractie zecimala nr x+y ; y-x .

    \rm{x=\frac{29}{10^3}=\frac{29}{1000}=0,029\\y=\frac{42}{10^2}=\frac{42}{100}=0,42.\bl}

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