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a ) \(VT=\left(x+y+z\right)^2-\left(x-y-z\right)^2\)
\(=\left(x+y+z-x+y+z\right)\left(x+y+z+x-y-z\right)\)
\(=4x\left(y+z\right)=VP\)
b ) \(VT=\left(2a+b\right)^2-\left(a+b\right)^2-3a^2\)
\(=\left(2a+b-a-b\right)\left(2a+b+a+b\right)-3a^2\)
\(=a\left(3a+2b\right)-3a^2\)
\(=3a^2+2ab-3a^2=2ab=VP\)
a) \(\left(x+y+z\right)^2-\left(x-y-z\right)^2=4x\left(y+z\right)\)
\(\Rightarrow x^2+y^2+z^2+2xy+2xz+2yz-\left(x^2+y^2+z^2-2xy-2xz+2yz\right)=4x\left(y+z\right)\)\(\Rightarrow x^2+y^2+z^2+2xy+2xz+2yz-x^2-y^2-z^2+2xy+2xz-2yz=4x\left(y+z\right)\)\(\Leftrightarrow4xy+4xz=4x\left(y+z\right)\)
\(\Leftrightarrow4x\left(y+z\right)=4x\left(y+z\right)\).
b) \(\left(2a+b\right)^2-\left(a+b\right)^2-3a^2=2ab\)
\(\Rightarrow\left(2a\right)^2+2.2a.b+b^2-\left(a^2+2ab+b^2\right)-3a^2=2ab\)
\(\Rightarrow4a^2+4ab+b^2-a^2-2ab-b^2-3a^2=2ab\)
\(\Leftrightarrow2ab=2ab\)
A2 + 2AB + B2 = ( A + B )2
A2 - 2AB + B2 = ( A - B ) 2
( A - B ) ( A + B ) = A2 - B2
~ Hok tốt ~
hằng đẳng thức thứ nhất sai rồi bạn , phải là
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
Ta có:
\(VP=4p\left(p-a\right)=2p.2p-2a.2p\)(1)
Thay \(a+b+c=2p\) vào (1) ta có:
\(\left(a+b+c\right)^2-2a.\left(a+b+c\right)\)
\(=a^2+b^2+c^2+2ab+2ac+2bc-2a^2-2ab-2ac\)
\(=-a^2+b^2+c^2+2bc=VT\)
Vậy \(2ab+b^2+c^2-a^2=4p\left(p-a\right)\)(đpcm)
Chúc bạn học tốt!!!
Ta có:a+b+c=2p=>b+c=2p-a=>b+c-a=2p-2a
Ta lại có:4p(p-a)=2p(2p-2a)=2(a+b+c)(b+c-a)=ab+ac-a2+b2+bc-ab+bc+c2-ac
=2ab+b2+c2-a2(đpcm)
a,\(x^2+2ab+b^2=\left(a+b\right)^2\)
b,\(x^2-2ab+b^2=\left(a-b\right)^2\)
(a+b)2=(a+b)(a+b)
=a2+ab+ab+b2
=a2+2ab+b2
Vậy (a+b)2=a2+2ab+b2