Cho a,b,c ∈ R và ( $(a+b+c)(\frac{1}{a}+$$\frac{1}{b}+$$\frac{1}{c})$ Tính P= $(a^{2004}-$ $b^{2004})($ $b^{2005}+$ $c^{2005})($ $c^{2006}-$ $a^

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Cho a,b,c ∈ R và ( $(a+b+c)(\frac{1}{a}+$$\frac{1}{b}+$$\frac{1}{c})$
Tính P= $(a^{2004}-$ $b^{2004})($ $b^{2005}+$ $c^{2005})($ $c^{2006}-$ $a^{2006})$

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  1. Đáp án: $P=0$

     

    Giải thích các bước giải:

    Từ `(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=1`

    `⇔\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}`

    `⇔\frac{ab+bc+ac}{abc}=\frac{1}{a+b+c}`

    $⇔(ab+bc+ac)(a+b+c)=abc$

    $⇔a^2b+ab^2+abc+abc+b^2c+bc^2+a^2c+abc+ac^2=abc$

    $⇔a^2b+ab^2+b^2c+bc^2+a^2c+ac^2+2abc=0$

    $⇔(a^2b+ab^2+b^2c+abc)+(bc^2+a^2c+ac^2+abc)=0$

    $⇔b(a^2+ab+bc+ac)+c(bc+a^2+ac+ab)=0$

    $⇔(b+c)(a^2+ab+bc+ac)=0$

    $⇔(b+c)[a(a+b)+c(a+b)]=0$

    $⇔(b+c)(a+c)(a+b)=0$

    $⇔\left[ \begin{array}{l}b+c=0\\a+c=0\\a+b=0\end{array} \right.$

    $⇔\left[ \begin{array}{l}b=-c\\a=-c\\a=-b\end{array} \right.$

    -Nếu $a=-b⇒a^{2004}-b^{2004}=(-b)^{2004}-b^{2004}=0$

    Khi đó: $P=0.(b^{2005}+c^{2005})(c^{2006}-a^{2006})=0$

    -Nếu $a=-c⇒c^{2006}-a^{2006}=c^{2006}-(-c)^{2006}=0$

    Khi đó: $P=(a^{2004}-b^{2004}).(b^{2005}+c^{2005}).0=0$

    -Nếu $b=-c⇒b^{2005}+c^{2005}=(-c)^{2005}+c^{2005}=0$

    Khi đó: $P=(a^{2004}-b^{2004}).0.(c^{2006}-a^{2006})=0$

    Vậy $P=0$

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