Bài 4: Phương trình tích

NT

1. Thực hiện phép tính:

\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)

ND
14 tháng 12 2017 lúc 20:59

\(\dfrac{1}{a\left(a-b\right)\left(a-c\right)}+\dfrac{1}{b\left(b-a\right)\left(b-c\right)}+\dfrac{1}{c\left(c-a\right)\left(c-b\right)}\)

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

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

\(=\dfrac{b^2c-bc^2-a^{ 2}c+ac^2+a^2b-ab^2}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

\(=\dfrac{a^2\left(b-c\right)-b^2\left[\left(b-c\right)+\left(a-b\right)\right]+c^2\left(a-b\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{\left(b-c\right)\left(a^2-b^2\right)-\left(a-b\right)\left(b^2-c^2\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

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

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

\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{abc\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\dfrac{1}{abc}\)

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