a+b +c = 2p 

 =>  b +c = 2p - a

=>  ( b + c)^2  = ( 2p -a)^2

=> b^2 + 2bc + c^2 = 4p^2 - 4ap + a^2

=> 2bc + b^2 + c^2 - a^2 = 4p^2 - 4ap

=> 2bc + b^2 + c^2 - a^2 = 4p ( p-a) 

=> ĐPCM 

( Xem lại đè = 4p(p - a) chứ không phải 4b( p-a)

a + b +c = 2p => b + c = 2p- a 

=> ( b + c)^2 = ( 2p -a )^2 

=> b^2 + 2bc + c^2 = 4p^2 - 4pa +a^2

=> b^2 + 2bc + c^2 - a^2 = 4p^2 - 4pa 

=> 2bc + b^2 + c^2 - a^2 = 4p(p-a) 

=> ĐPCM

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

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

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

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

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

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

\(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)

Ta có:VT=\(\left(b+c\right)^2-a^2=\)\(\left(b+c-a\right)\left(a+b+c\right)=2p\left(2p-2a\right)\)

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

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

Mộc Lung Hoađề bài

a + b +c = 2P => b+ c = 2P -a 

=> ( b +c )^2 =( 2P -a )^ 2 => b^2 +c^2 +2bc = 4P^2 - 4Pa + a^2

      = 2bc +  b^2 +c^2 - a^2 = 4P( P -a ) => ĐPCM

4p(p-a)=2p(2p-2a)=(a+b+c)(b+c-a)=-a^2+b^2+2bc+c^2=VT=>đpcm

Ta có: 2bc+b²+c²-a²=(b²+2bc+c²)-a²

                             =(b+c)²-a² (1)

Mà: a+b+c=2p=> b+c=2p-a. Thay b+c=2p vào (1) ta có:

               (2p-a)²-a²=4p²-4ap+a²-a²=4p²-4ap=4p.(p-a) (ĐPCM)

Gọi  \(2bc+b^2 +c^2-a^2=VT\)

và \(4p\left(p-a\right)=VP\)

Biến đổi VP ta có :

\(4p\left(p-a\right)=2p\left(2p-2a\right)\)

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

\(=2bc+b^2+c^2-a^2=VT\)  (đpcm)

Vậy ......

Ta có: \(a+b+c=2p\)

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

\(\Rightarrow b^2+2bc+c^2=4p^2-4pa+a^2\)

\(\Rightarrow2bc+b^2+c^2-a^2=4p\left(p-a\right)\)(đpcm)

Vậy....

Ta có :

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

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

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

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

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

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

\(\left(đpcm\right)\)