Question

chứng minh rằng

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

Answer 1

Lời giải:

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

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

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

Xét \(bc(c-b)+ac(a-c)+ab(b-a)=bc(c-b)-ac[(c-b)+(b-a)]+ab(b-a)\)

\(=(c-b)(bc-ac)+(b-a)(ab-ac)=c(c-b)(b-a)+a(b-a)(b-c)\)

\(=(c-b)(b-a)(c-a)=(a-b)(b-c)(c-a)\) (2)

Từ \((1),(2)\Rightarrow \text{VT}=\frac{(a-b)(b-c)(c-a)}{abc(a-b)(b-c)(c-a)}=\frac{1}{abc}\)

Ta có đpcm.