Question

Bai 1. Phan tich da thuc thanh nhan tu

x² +4x-y²+4

Bai 2 Cho a+b+c=0.Chung minh a³+b³+c³ =  3abc

Answer 1

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 \(a+b=-c\) )

\(\Rightarrow a^3+b^3+c^3=3abc\)

Answer 2

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)