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![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](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 + c^2 + 2bc = 4p^2 - 4pa + a^2
=> b^2 + c^2 + 2bc - a^2 = 4p(p-a)
=> 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....
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2bc+b^2+c^2-a^2\)
\(=\left(b+c\right)^2-a^2\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=2p\left(a+b+c-2a\right)\)
\(=2p\left(2p-2a\right)=4p\left(p-a\right)\)
biến đổi vế phải ta được:
4p(p -a ) = 4p\(^2\)-4pa
=(2p)\(^2\)-2p.2a
=(a+b+c)\(^2\)-2a(a+b+c)
=\(a^2+b^2+c^2+2ab+2ac+2bc\)-\(2a^2-2ab-2ac\)
=\(2bc+b^2+c^2-a^2\)=vế trái (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2bc+b^2+c^2-a^2.\)'
\(=\left(2bc+b^2+c^2\right)-a^2.\)
\(=\left(b+c\right)^2-a^2\)
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 )
![](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)
![](https://rs.olm.vn/images/avt/0.png?1311)
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
4:
a: 2003*2005=(2004-1)(2004+1)=2004^2-1<2004^2
b: 8(7^2+1)(7^4+1)(7^8+1)
=1/6*(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^2-1)(7^2+1)(7^4+1)(7^8+1)
=1/6(7^16-1)<7^16-1
5:
a: (2x-5)(2x+5)=4x^2-25
b: (3x-5y)(3x+5y)=9x^2-25y^2
c: (3x+7y)(3x-7y)=9x^2-49y^2
d: (2x-1)(2x+1)=4x^2-1
mik chỉ biết bài 5 thôi !