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\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow abc.\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\Leftrightarrow\hept{\begin{cases}bc=-\left(ab+ac\right)\\ab=-\left(bc+ac\right)\\ac=-\left(bc+ab\right)\end{cases}}\)
Ta có: \(a^2+2bc=a^2+bc+bc=a^2+bc+\left(-ab-ac\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự \(b^2+2ac=\left(b-a\right)\left(b-c\right);c^2+2ab=\left(c-a\right)\left(c-b\right)\)
\(\Leftrightarrow N=\frac{bc}{\left(a-b\right)\left(a-c\right)}+\frac{ac}{\left(b-a\right)\left(b-c\right)}+\frac{ab}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a^2-b^2\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)
Ta có:\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-bc-ac\\bc=-ac-ab\\ac=-ab-bc\end{cases}}\)(*)
Thay (*) vào M ta được:
\(M=\frac{1}{a^2+bc-ab-ac}+\frac{1}{b^2+ac-ab-bc}+\frac{1}{c^2+ab-bc-ac}\)
\(=\frac{1}{a\left(a-b\right)-c\left(a-b\right)}+\frac{1}{a\left(c-b\right)-b\left(c-b\right)}+\frac{1}{c\left(c-a\right)-b\left(c-a\right)}\)
\(=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(a-b\right)\left(c-b\right)}-\frac{1}{\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}+\frac{a-c}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}-\frac{a-b}{\left(a-b\right)\left(c-b\right)\left(a-c\right)}\)
\(=\frac{c-b+a-c-a+b}{\left(a-b\right)\left(a-c\right)\left(c-b\right)}=0\)
Vậy M = 0
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Rightarrow ab+bc+ac=0\Rightarrow\hept{\begin{cases}ab=-ac-bc\\bc=-ab-ac\\ac=-ab-bc\end{cases}}\)
\(a^2+2bc=a^2+bc+bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự: \(b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(c^2+2ab=\left(a-c\right)\left(b-c\right)\)
\(B=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}+\frac{ca+1}{\left(b-a\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}-\frac{ca+1}{\left(a-b\right)\left(b-c\right)}+\frac{ab+1}{\left(a-c\right)\left(b-c\right)}\)
\(=\frac{\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-c\right)+\left(ab+1\right)\left(a-b\right)\)
\(=\left(bc+1\right)\left(b-c\right)-\left(ca+1\right)\left(a-b\right)-\left(ca+1\right)\left(b-c\right)+\left(ab+1\right)\left(a-b\right)\)
\(=\left(b-c\right)\left(bc+1-ca-1\right)+\left(a-b\right)\left(ab+1-ca-1\right)\)
\(=\left(b-c\right)\left(bc-ca\right)+\left(a-b\right)\left(ab-ca\right)\)
\(=\left(b-c\right)c\left(b-a\right)+\left(a-b\right)a\left(b-c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Vậy B = 1
Từ gt => ab+bc+ca=0
\(a^2+2bc=a^2+bc+\left(-ab-ac\right)=a\left(a-b\right)-c\left(a-b\right)=\left(a-b\right)\left(a-c\right)\)
Tương tự \(\hept{\begin{cases}b^2+2ac=\left(b-a\right)\left(b-c\right)\\c^2+2ab=\left(c-a\right)\left(c-b\right)\end{cases}}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}=\frac{b-c+c-a+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow ab+bc+ca=0\)
\(\Rightarrow bc=-ab-ca\)
Vậy thì \(a^2+2bc=a^2+bc-ab-ac=a\left(a-b\right)-c\left(a-b\right)=\left(a-c\right)\left(a-b\right)\)
Tương tự ta cũng có:
\(b^2+2ac=\left(b-c\right)\left(b-a\right)\)
\(c^2+2ab=\left(c-a\right)\left(c-b\right)\)
Vậy thì \(A=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(A=\frac{-b+c+a-c-a+b}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(A=0.\)