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Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)
\(A=\frac{2}{x}+\frac{2}{y}+\frac{2}{z}+\frac{x^2y^2z^2}{xyz}\)
\(A=\frac{\left(2y+2x\right).z+2xy}{xyz}+\frac{x^2+y^2+x^2}{xyz}\)
\(A=\frac{2yz+2xz+2xy}{xyz}+\frac{x^2+y^2+z^2}{xyz}\)
\(A=\frac{2yz+2xz+2xy+x^2+y^2+z^2}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}\)
Có đúng k nhỉ k chắc
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
MTC: \(abc\left(a-b\right)\left(b-c\right)\left(a-c\right)\)nên
\(A=\frac{bc\left(b-c\right)\left(a-2\right)\left(a-1014\right)}{abc\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{ac\left(a-c\right)\left(b-2\right)\left(b-1004\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}+\frac{ab\left(a-b\right)\left(c-2\right)\left(c-1004\right)}{abc\left(a-c\right)\left(a-b\right)\left(b-c\right)}\)
\(=\frac{2008b^2c+2008a^2c+2008a^2b-2008bc^2-2008a^2c-2008ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left[\left(c^2a-c^2b\right)+\left(a^2b-a^2c\right)+\left(b^2a-b^2c\right)\right]}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)
\(=\frac{2008}{abc}\) ( với \(abc\ne0\))