Question

[Nguyễn Tuấn Minh]

Cho a,b,c đôi một khác nhau và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

Tính \(A=\frac{bc+1}{a^2+2bc}+\frac{ac+1}{b^2+2ac}+\frac{ab+1}{c^2+2ab}\)

Answer 1

[Không Tên]

    \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)

\(\Leftrightarrow\)\(\frac{bc+ac+ab}{abc}=0\)

\(\Leftrightarrow\)\(bc+ac+ab=0\)

\(\Leftrightarrow\)\(\hept{\begin{cases}bc=-ab-ac\\ac=-ab-bc\\ab=-bc-ac\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}a^2+2bc=a^2+bc-ab-ac=\left(a-b\right)\left(a-c\right)\\b^2+2ac=b^2+ac-ab-bc=\left(b-a\right)\left(b-c\right)\\c^2+2ab=c^2+ab-bc-ac=\left(c-a\right)\left(c-b\right)\end{cases}}\)

\(A=\frac{bc+1}{\left(a-b\right)\left(a-c\right)}+\frac{ac+1}{\left(b-a\right)\left(b-c\right)}+\frac{ab+1}{\left(c-a\right)\left(c-b\right)}\)

= \(\frac{bc\left(b-c\right)+b-c+ac\left(c-a\right)+c-a+ab\left(a-b\right)+a-b}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{bc\left(b-c\right)+ca\left(c-a\right)-ab\left(b-c\right)-ab\left(c-a\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{\left(b-c\right)\left(bc-ab\right)+\left(c-a\right)\left(ca-ab\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{b\left(b-c\right)\left(c-a\right)+a\left(c-a\right)\left(c-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

= \(\frac{\left(a-c\right)\left(b-c\right)\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=1\)