Question

Cho a+b+c=2p.Chứng minh hằng đẳng thức:

2ab+b²+c²-a²=4p(p-a)

Answer 1

Ta có:

\(VP=4p\left(p-a\right)=2p.2p-2a.2p\)(1)

Thay \(a+b+c=2p\) vào (1) ta có:

\(\left(a+b+c\right)^2-2a.\left(a+b+c\right)\)

\(=a^2+b^2+c^2+2ab+2ac+2bc-2a^2-2ab-2ac\)

\(=-a^2+b^2+c^2+2bc=VT\)

Vậy \(2ab+b^2+c^2-a^2=4p\left(p-a\right)\)(đpcm)

Chúc bạn học tốt!!!

Answer 2

Ta có:a+b+c=2p=>b+c=2p-a=>b+c-a=2p-2a

Ta lại có:4p(p-a)=2p(2p-2a)=2(a+b+c)(b+c-a)=ab+ac-a²+b²+bc-ab+bc+c²-ac

=2ab+b²+c²-a²(đpcm)

Answer 3

Ta có: a+b+c = 2p thì:

2ab+b²+c²-a² = 4p(p-a)

<=> 2ab+b²+c²-a² = 2p(2p-2a)

<=> 2ab+b²+c²-a² = (a+b+c)(b+c-a)

<=> 2ab+b²+c²-a² = ab+ac+(-a)²+b²+bc-ab+bc+c²-ac

<=> 2ab+b²+c²-a² = 2ab+b²+c²-a²

Vây: 2ab+b²+c²-a² = 4p(p-a)