1) CMR nếu $a+b+c=0$ thì $a^3+b^3+c^3=3abc$?
CMR nếu $a+b+c=0$ thì $a^3+b^3+c^3=3abc$?
#1
Posted 30-04-2014 - 20:56
#2
Posted 30-04-2014 - 21:11
$a^3+b^3+c^3=3abc$
$\Leftrightarrow (a^3+b^3)+c^3-3abc=0$
$\Leftrightarrow (a+b)(a^2-ab+b^2)+c^3-3abc=0$
$\Leftrightarrow -c(a^2-ab+b^2)+c^3-3abc=0$
$\Leftrightarrow c(-a^2+ab-b^2+c^2-3ab)=0$
$\Leftrightarrow c\left [ -(a+b)^2+c^2 \right ]=0$
$\Leftrightarrow c(-c^2+c^2)=0$ (Luôn đúng)
Bài toán được chứng minh!
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#3
Posted 30-04-2014 - 21:19
Vì $a+b+c=0\Rightarrow a+b=-c$. Ta có:
$(a+b)^{3}=(-c)^{3}\Rightarrow (a+b)^{3}+c^{3}=0\Leftrightarrow a^{3}+b^{3}+c^{3}+3ab(a+b)=0\Leftrightarrow a^{3}+b^{3}+c^{3}-3abc=0$ (vì $a+b=-c$) $\Rightarrow a^{3}+b^{3}+c^{3}=3abc$.
Edited by SuperReshiram, 30-04-2014 - 21:21.
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#4
Posted 16-05-2014 - 12:51
Xét $a^{3}+b^{3}+c^{3}-abc= \frac{1}{2}\times \left ( a+b+c \right )\left [ \left ( a-b \right )^{2}+ \left ( b-c \right )^{2}+\left ( c-a \right )^{2}\right)
màa+b+c=0$\Rightarrow$điều phải chứng minh
- sjeunhan1998 likes this
#5
Posted 22-08-2015 - 22:24
$a^3+b^3+c^3=3abc$
$\Leftrightarrow (a^3+b^3)+c^3-3abc=0$
$\Leftrightarrow (a+b)(a^2-ab+b^2)+c^3-3abc=0$
$\Leftrightarrow -c(a^2-ab+b^2)+c^3-3abc=0$
$\Leftrightarrow c(-a^2+ab-b^2+c^2-3ab)=0$
$\Leftrightarrow c\left [ -(a+b)^2+c^2 \right ]=0$
$\Leftrightarrow c(-c^2+c^2)=0$ (Luôn đúng)
Bài toán được chứng minh!
Edited by anhtukhon1, 22-08-2015 - 22:26.
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