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!!!
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)
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)
