[Toán 8] toán Chứng minh

S

serena_tsukino

T

transformers123

a/ $a^3+b^3+c^3=3abc$

$\iff (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$

$\bigstar TH_1:\ a+b+c=0$

$\bigstar TH_2:\ a^2+b^2+c^2-ab-bc-ca=0$

$\iff \dfrac{1}{2}[(a-b)^2+(b-c)^2+(c-a)^2]=0$

$\iff \begin{cases}a-b=0\\b-c=0\\c-a=0\end{cases} \iff a=b=c$

b/ Xét $a+b+c=0$, ta có:

$M=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$

$\iff M=\dfrac{a}{-a}+\dfrac{b}{-b}+\dfrac{c}{-c}$

$\iff M=-3$

Xét $a=b=c$, ta có:

$M=\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$

$\iff M=\dfrac{a}{2a}+\dfrac{b}{2b}+\dfrac{c}{2c}$

$\iff M=\dfrac{3}{2}$
 
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