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Pe: 2021-01-24T23:53:41+02:00 In: In:

Buna seara! Oare are cineva idee cum …

Buna seara! Oare are cineva idee cum s-ar rezolva acesti determinanți?

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  1. maestru (V)

    a)\begin{vmatrix} a+b&b+c &c+a \\ a^2+b^2&b^2+c^2 &c^2+a^2 \\ a^3+b^3&b^3+c^3 &c^3+a^3 \end{vmatrix}=2abc(a-b)(a-c)(b-c)
    Scadem coloana 2 din coloana 1 si coloana 3 din coloana 2:
    \begin{vmatrix} a-c &b-a &c+a \\ a^2-c^2&b^2-a^2 &c^2+a^2 \\ a^3-c^3&b^3-a^3 &c^3+a^3 \end{vmatrix}=2abc(a-b)(a-c)(b-c)
    \begin{vmatrix} a-c &b-a &c+a \\ (a-c)(a+c)&(b-a)(b+a) &c^2+a^2 \\ (a-c)(a^2+ac+c^2)&(b-a)(b^2+ab+a^2) & c^3+a^3 \end{vmatrix}=2abc(a-b)(a-c)(b-c)
    Scoatem factori comuni pe primele 2 coloane:
    (a-c)(b-a)\begin{vmatrix} 1 &1 &c+a \\ (a+c)&(b+a) &c^2+a^2 \\ (a^2+ac+c^2)&(b^2+ab+a^2) & c^3+a^3 \end{vmatrix}=2abc(a-b)(a-c)(b-c)
    Eliminam factorii comuni:
    \begin{vmatrix} 1 &1 &c+a \\ (a+c)&(b+a) &c^2+a^2 \\ (a^2+ac+c^2)&(b^2+ab+a^2) & c^3+a^3 \end{vmatrix}=-2abc(b-c)
    Scadem din nou a 2-a coloana din prima:
    \begin{vmatrix} 0 &1 &c+a \\ c-b&(b+a) &c^2+a^2 \\ c^2-b^2+a(c-b)&(b^2+ab+a^2) & c^3+a^3 \end{vmatrix}=-2abc(b-c)
    Scoatem factor comun pe c-b. Eliminam acest factor:
    \begin{vmatrix} 0 &1 &c+a \\ 1 &(b+a) &c^2+a^2 \\ c+b+a&(b^2+ab+a^2) & c^3+a^3 \end{vmatrix}=2abc
    Dezvoltand dupa prima linie:
    0((b+a)(c^3+a^3)-(b^2+ab+a^2)(c^2+a^2))-(c^3+a^3-(c^2+a^2)(a+b+c))+(c+a)(b^2+ab+a^2-(a+b+c)(a+b))=2abc
    -(c^3+a^3-ac^2-bc^2-c^3-a^3-a^2b-a^2c)+(c+a)(b^2+ab+a^2-a^2-ab-ab-b^2-ac-bc)=2abc
    -(-ac^2-bc^2-a^2b-a^2c)+(c+a)(-ab-ac-bc)=2abc
    ac^2+bc^2+a^2b+a^2c-abc-ac^2-bc^2-a^2b-a^2c-abc=2abc
    -2abc=2abc
    Egalitatea data nu pare sa fie adevarata, daca nu am gresit eu nimic in calcule. Voi verifica acest subpunct mai tarziu.

    b)Adunam toate liniile peste prima:
    \begin{vmatrix} a+b+c&a+b+c&a+b+c \\ 2b&b-c-a &2b \\ 2c&2c &c-a-b \end{vmatrix}=(a+b+c)^3
    Simplificam cu a+b+c-ul dat de prima linie:
    \begin{vmatrix} 1&1&1 \\ 2b&b-c-a &2b \\ 2c&2c &c-a-b \end{vmatrix}=(a+b+c)^2
    Scadem a 3-a coloana din prima:
    \begin{vmatrix} 0& 1 & 1\\ 0& b-c-a & 2b\\ a+b+c& 2c& c-a-b \end{vmatrix}=(a+b+c)^2
    Simplificam cu factorul comun a+b+c dat de prima coloana:
    \begin{vmatrix} 0& 1 & 1\\ 0& b-c-a & 2b\\ 1& 2c& c-a-b \end{vmatrix}=a+b+c
    Dezvoltand dupa prima coloana:
    2b-(b-c-a)=(a+b+c)
    Aceasta ultima egalitate este adevarata.

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