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ta có : a+b+c=0=>a+b=-c ; b+c=-a ; a+c=-b
ta có: M= \(\frac{2ab}{a^2+\left(b+c\right)\left(b-c\right)}+\frac{2bc}{b^2+\left(c+a\right)\left(c-a\right)}+\frac{2ca}{c^2+\left(a+b\right)\left(a-b\right)}\)
M=\(\frac{2ab}{a^2-a\left(b-c\right)}+\frac{2bc}{b^2-b\left(c-a\right)}+\frac{2ca}{c^2-c\left(a-b\right)}\)
M=\(\frac{2ab}{a\left(a-b+c\right)}+\frac{2bc}{b\left(b-c+a\right)}+\frac{2ca}{c\left(c-a+b\right)}\)
M=\(\frac{2ab}{-ab+\left(a+c\right)}+\frac{2bc}{-bc+\left(a+b\right)}+\frac{2ac}{-ac+\left(b+c\right)}\)
M=\(\frac{2ab}{-2ab}+\frac{2bc}{-2bc}+\frac{2ca}{-2ca}\)
M=-1-1-1=-3
Vậy với a+b+c=0 thì M=-3
A \(=\frac{2}{a-b}+\frac{2}{b-c}+\frac{2}{c-a}+\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\)
\(=\frac{2\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{2\left(a-b\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{2\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}+\frac{\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\)
\(=\frac{2\left(b-c\right)\left(c-a\right)+2\left(a-b\right)\left(c-a\right)+2\left(a-b\right)\left(b-c\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\)
\(=\frac{2ab+2ac+2bc-2a^2-2b^2-2c^2+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\)
\(=\frac{-\left(a^2-2ab+b^2\right)-\left(b^2-2bc+c^2\right)-\left(c^2-2ac+a^2\right)+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\)
\(=\frac{-\left(a-b\right)^2-\left(b-c\right)^2-\left(c-a\right)^2+\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\)
\(=\frac{0}{\left(a-b\right)\left(b-a\right)\left(c-a\right)}\) = 0
Ta có
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}=\frac{a^2+ab-bc-ab}{\left(a+b\right)\left(a+c\right)}=\frac{a\cdot\left(a+b\right)-b\cdot\left(c+a\right)}{\left(a+b\right)\left(c+a\right)}=\frac{a}{a+c}-\frac{b}{a+b}\left(1\right)\)
tương tự
\(\frac{b^2-bc}{\left(a+b\right)\left(b+c\right)}=\frac{b}{a+b}-\frac{c}{b+c}\left(2\right)\)
\(\frac{c^2-ab}{\left(c+a\right)\left(b+c\right)}=\frac{c}{c+b}-\frac{a}{a+b}\left(3\right)\)
Cộng (1);(2) và (3) ta có
\(\frac{a^2-bc}{\left(a+b\right)\left(a+c\right)}+\frac{b^2-ac}{\left(a+b\right)\left(b+c\right)}+\frac{c^2-ab}{\left(a+c\right)\left(c+b\right)}=\frac{a}{a+c}-\frac{b}{a+b}+\frac{b}{a+b}-\frac{c}{b+c}+\frac{c}{c+b}-\frac{a}{a+b}=0 \)