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17 tháng 11 2019

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 \(abc\ne0\))

13 tháng 10 2019

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

2 tháng 11 2019

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

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

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

=> A + \(\frac{ab-ac-ab+bc+ac-bc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\)

2 tháng 11 2019

\(B=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-a\right)\left(b-c\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(b-a\right)\left(b-c\right)\left(c-a\right)}\)

\(+\frac{c^2\left(a-b\right)}{\left(c-a\right)\left(c-b\right)\left(a-b\right)}\)

\(=\frac{a^2\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{b^2\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(+\frac{c^2\left(a-b\right)}{\left(a-c\right)\left(b-c\right)\left(a-b\right)}\)

\(=\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

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

4 tháng 4 2017

thiếu đề bài òi bạn ko làm đc đâu

7 tháng 4 2017

Mik giải ra rồi!