Bài 1 :
\(x^2+4x-y^2+4\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2+y\right)\left(x+2-y\right)\)
Bài 2 : Ta có : \(a+b+c=0\)
\(\Rightarrow a+b=-c\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3-3abc=-c^3\) ( Vì \(a+b=-c\) )
\(\Rightarrow a^3+b^3+c^3=3abc\)
Bài 1:
x2 +4x-y2+4
=(x2+4x+4)-y2
=(x+2)2-y2
=(x-y+2)(x+y+2)
Bài 2:
a3+b3+c3 = 3abc
=>a3+b3+c3-3abc=0
=>[(a+b)3+c3]-3ab(a+b)-3abc=0
=>(a+b+c)[(a+b)2-(a+b)c+c2]-3ab(a+b+c)=0
=>(a+b+c)(a2+b2+c2-ac-bc-ab)=0
Từ a+b+c=0
=>0*(a2+b2+c2-ac-bc-ab)=0 (luôn đúng)
