簡約 1/((a-b)(a-c))+1/((b-c)(b-a))+1/((c-a)(c-b))

解

簡約

\frac{1}{( a - b ) ( a - c )} + \frac{1}{( b - c ) ( b - a )} + \frac{1}{( c - a ) ( c - b )}

0

解答ステップ

\frac{1}{( a - b ) ( a - c )} + \frac{1}{( b - c ) ( b - a )} + \frac{1}{( c - a ) ( c - b )}

以下の最小公倍数:

(a - b) (a - c), (b - c) (b - a), (c - a) (c - b) : - (a - b) (a - c) (b - c)

LCMに基づいて分数を調整する

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

分数の規則を適用する:

\frac{a}{c} + \frac{b}{c} = \frac{a + b}{c}

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

簡素化

- b + c - c + a - a + b : 0

= \frac{0}{- ( a - b ) ( a - c ) ( b - c )}

規則を適用

\frac{0}{a} = 0

a \neq = 0

= 0