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\(a+b+c=2p\Rightarrow a=2p-b-c\)
Ta có:
\(a^2-b^2-c^2+2bc=a^2-\left(b-c\right)^2=\left(a-b+c\right)\left(a+b-c\right)\)
\(=\left(2p-b-c-b+c\right)\left(2p-b-c+b-c\right)\)
\(=\left(2p-2b\right)\left(2p-2c\right)\)
\(=4\left(p-b\right)\left(p-c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
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\(2bc+b^2+c^2-a^2\)
\(=\left(b+c\right)^2-a^2\)
\(=\left(b+c+a\right)\cdot\left(b+c-a\right)\)
\(=2p\cdot\left(2p-a-a\right)\)
\(=4p\left(p-a\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
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)
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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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ......
![](https://rs.olm.vn/images/avt/0.png?1311)
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: 4(p-c)(p-b)=(2p-2c)(2p-2b)=(a+b-c)(a+c-b)=[a+(b-c)].[a-(b-c)]=a^2 -(b-c)^2=a^2-b^2-c^2+2bc
1,a²-(b²+c²-2bc) = a² - (b-c)² = (a-b+c)(a+b-c)
=(a+b+c-2b)(a+b+c-2c) = (2p-2b)(2p-2c)=4(p-b)(p-c)
2,p²+(p-a)²+(p-b)²+(p-c)² = 4p² + (a²+b²+c²) - 2p(a+b+c)
= (a+b+c)² + (a²+b²+c²) - (a+b+c)² = (a²+b²+c²)