Vì:

a+b+c=2p => b+c=2p-a

Ta có (2bc+b^2+c^2)-a^2

= ( b+c)^2 -a^2 

= (2p-a)^2 - a^2

= 4p^2 - 4pa + a^2 -a^2

= 4p(p+a) => đpcm

k cho mình

Có 2p=a+b+c

Suy ra:4p(p-a)=2p(2p-2a)

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

=(a+b+c)(b+c-a)

=ab+ac-a^2+B^2+bc-ab+cb+c^2-ac

=2bc+b^2+c^2-a^2

    Nhớ nhé!

Cảm ơn

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)

\(a+b+c=2p\Rightarrow a=2p-b-c\)

Ta có:

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

\(=\left(2p-b-c-b+c\right)\left(2p-b-c+b-c\right)\)

\(=\left(2p-2b\right)\left(2p-2c\right)\)

\(=4\left(p-b\right)\left(p-c\right)\)

lớp 9 đấy

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

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

\(=\left(b+c+a\right)\cdot\left(b+c-a\right)\)

\(=2p\cdot\left(2p-a-a\right)\)

\(=4p\left(p-a\right)\)