Cho a+b+c = 2p . Chứng minh rằng đẳng thức : \(2bc+b^2+c^2-a^2=4p\left(p-a\right)\)
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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....
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 : ........
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
\(2bc+b^2+c^2-a^2\)
\(=\left(b+c\right)^2-a^2\)
\(=\left(a+b+c\right)\left(b+c-a\right)\)
\(=2p\left(a+b+c-2a\right)\)
\(=2p\left(2p-2a\right)=4p\left(p-a\right)\)
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)