Question

Biết a+b+c=2p.CMR:\(\dfrac{1}{p-a}+\dfrac{1}{p-b}+\dfrac{1}{p-c}-\dfrac{1}{p}=\dfrac{abc}{p\left(p-a\right)\left(p-b\right)\left(p-c\right)}\)

Answer 1

Lời giải:

Ta có:

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

\(=\frac{2p-(a+b)}{(p-a)(p-b)}+\frac{c}{p(p-c)}=\frac{(a+b+c)-(a+b)}{(p-a)(p-b)}+\frac{c}{p(p-c)}\)

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

\(=c.\frac{p(p-c)+(p-a)(p-b)}{p(p-a)(p-b)(p-c)}=c.\frac{2p^2-p(a+b+c)+ab}{p(p-a)(p-b)(p-c)}\)

\(=c.\frac{2p^2-p.2p+ab}{p(p-a)(p-b)(p-c)}=\frac{abc}{p(p-a)(p-b)(p-c)}\)

Ta có đpcm.