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 ......

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

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

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

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

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

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

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

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

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

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

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

\(\left(đpcm\right)\)

a+b+c = 2p => 4p = 2(a+b+c); p=(a+b+c)/2

VP = 4p(p-a) = 2(a+b+c)(\(\frac{a+b+c}{2}-a\))

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

=\(2\left(a+b+c\right)\cdot\frac{b+c-a}{2}=\left(a+b+c\right)\left(b+c-a\right)\)

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

\(=2bc+b^2+c^2-a^2\) = VT (đpcm)

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

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

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

Theo đề ta có \(a+b+c=2p\)

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

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

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

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

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

\(=2p\cdot2\left(p-a\right)=4p\left(p-a\right)\)

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

2bc + b² + c² - a²

= ( b² + 2ab + c² ) - a²

= ( b + c )² - a²

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

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

Thế vào (*) ta được

( 2p - a - a )( 2p - a + a )

= ( 2p - 2a )2p

= 4p² - 4pa

= 4p( p - a ) ( đpcm )

Ta có: \(VT=2bc+b^2+c^2-a^2\)

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

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

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

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

\(=2p\left(-2a+2p\right)\) 9 ( Vì 2p - a = b + c )

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

\(\Rightarrowđpcm\)

Ta có : \(4p\left(p-a\right)=2\left(a+b+c\right)\left(\dfrac{a+b+c}{2}-a\right)\)

\(=2\left(a+b+c\right)\left(\dfrac{b+c-a}{2}\right)\)

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

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

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

Vậy : ........

Trần Hoàng Nghĩa

Tuấn Anh Phan Nguyễn

Nguyễn Huy Tú

Ace Legona

soyeon_Tiểubàng giải

Võ Đông Anh Tuấn

a + b + c = 2p

=> b + c = 2p - a 

=> ( b + c )^2 = (2p - a )^2

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

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

=> ĐPCM 

TC:a+b+cd=2p=>b+c=2p-a

=>(b+c)²=(2p-a)²

=>b²+2bc+c²=4p²-4pa+a²

=>b²+2bc+c²-a²=4p²-4pa

=>2bc+b²+c²-a²=4p(p-a) ĐPCM