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. \(\Rightarrow\) b + c = 2P - a
\(\Rightarrow\) \(\left(b+c\right)^2=\left(2P-a\right)^2\Rightarrow b^2+c^2+2bc=\)\(4P^2-4Pa+a^2\) \(=2bc+b^2+c^2-a^2=4P\left(P-a\right)\Rightarrowđpcm\)
\(a + b +c = 2P => b+ c = 2P -a \)
\(\Rightarrow\) \(( b +c )^2 =( 2P -a )^ 2 => b^2 +c^2 +2bc = 4P^2 - 4Pa + a^2\)
\(= 2bc + b^2 +c^2 - a^2 = 4P( P -a ) => ĐPCM\)
