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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 8: (a+b)(b+c)(a+c)=abc;(a3+b3)(b3+c3)(a3+c3)=a3b3c3.cm abc=0

Tháng Mười Một 10, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: (a+b)(b+c)(a+c)=abc;(a3+b3)(b3+c3)(a3+c3)=a3b3c3.cm abc=0

Comments ( 1 )

  1. aivilanlan
    aivilanlan
    10 Tháng Mười Một, 2022 at 2:31 sáng
    Reply
    $(a^{3}+b^3)(b^3+c^3)(c^3+a^3)=a^3b^3c^3$
    <=> $(a+b)(a^{2}-ab+b^2)(b+c)(b^2-bc+c^2)(a+c)(c^2-ac+a^2)=a^3b^3c^3$
    <=> $(a+b)(b+c)(a+c)(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)=a^3b^3c^3$
    <=> $abc(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)=a^3b^3c^3$
    <=> $abc(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)-a^3b^3c^3=0$
    <=> $abc\left[\begin{array}{ccc}(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)-a^2b^2c^2\\\end{array}\right]=0$
    <=> \(\left[ \begin{array}{l}abc=0(1)\\\left[\begin{array}{ccc}(a^{2}-ab+b^2)(b^2-bc+c^2)(c^2-ac+a^2)-a^2b^2c^2\\\end{array}\right]=0(2)\end{array} \right.\)
    <=> (2) $(a^{2}-ab+b^2)=a^2-2a.\dfrac{b}{2}+\dfrac{b^2}{4}+\dfrac{3b^2}{4}=(a-\dfrac{b}{2})^2+\dfrac{3b^2}{4}$
    Ta có : $(a-\dfrac{b}{2})^{2}\ge0(\forall a,b)$
    => $(a-\dfrac{b}{2})^{2}+\dfrac{3b^2}{4}\ge\dfrac{3b^2}{4}$
    $\text{Tương tự ta có:}$
    $(b-\dfrac{c}{2})^{2}+\dfrac{3c^2}{4}\ge\dfrac{3c^2}{4}$ $(c-\dfrac{a}{2})^{2}+\dfrac{3a^2}{4}\ge\dfrac{3a^2}{4}$
    Vậy $(a^2-ab+b^2)(b^2-bc+c^2)(c^2-ca+a^2)^{}=\dfrac{27}{64}a^2b^2c^2(3)$
    $\text{Từ (2) và(3) suy ra:}$
    => $a^{2}b^2c^2=0$
    => $abc=0(4)$ $\text{(1)(4) => abc=0 (đpcm)}$
    @K

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222-9+11+12:2*14+14 = ? ( )

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