Ta có : 1/M=a2+2bc+b2+2ac+c2+2ab
=(a+b+c)2 ➝ M=1/(a+b+c)2
mik nghĩ là thế
Có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)
\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\)
\(1\Leftrightarrow a^2+2bc=a^2+bc-ab-ac\)
\(\Leftrightarrow a^2+2bc=a\left(a-b\right)-c\left(a-b\right)\)
\(\Leftrightarrow a^2+2bc=\left(a-b\right)\left(b-c\right)\)
\(2\Leftrightarrow b^2+2ac=b^2+ac-ab-bc\)
\(\Leftrightarrow b^2+2ac=b\left(b-c\right)-a\left(b-c\right)\)
\(\Leftrightarrow b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(3.c^2+2ab=c^2+ab-bc-ac\)
\(\Leftrightarrow c^2+2ab=c\left(c-b\right)-a\left(c-b\right)\)
\(\Leftrightarrow c^2+2ab=\left(c-a\right)\left(c-b\right)\)
\(\Rightarrow M=\dfrac{1}{\left(a-b\right)\left(a-c\right)}+\dfrac{1}{\left(b-a\right)\left(b-c\right)}+\dfrac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\Rightarrow M=\dfrac{1}{\left(a-b\right)\left(a-c\right)}-\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(a-c\right)\left(b-c\right)}\)
\(\Rightarrow M=\dfrac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\Rightarrow M=0\)
