toán 8 hsg

K

kun_kun.chishj

Last edited by a moderator:
T

transformers123

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}$

$\iff \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0$

$\iff \dfrac{ab+bc+ca}{abc}-\dfrac{1}{a+b+c}=0$

$\iff \dfrac{(ab+bc+ca)(a+b+c)-abc}{abc(a+b+c)}=0$

$\Longrightarrow (ab+bc+ca)(a+b+c)-abc=0$

$\iff (a+b)(b+c)(c+a)=0$

$\iff a=-b$ hoặc $b=-c$ hoặc $c=-a$

Chia làm $3$ trường hợp, thế vào $M$ là xong :D
 
Last edited by a moderator:
Top Bottom