\(x+y+z=6\Rightarrow\left(x+y+z\right)^2=36\Rightarrow x^2+y^2+z^2+2\left(xy+yz+xz\right)=36\)

\(\Rightarrow xy+yz+zx=\frac{36-\left(x^2+y^2+z^2\right)}{2}=\frac{36-12}{2}=12=x^2+y^2+z^2\)(1)

Mặt khác ta luôn có: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)

hay: \(2x^2+2y^2+2z^2-2\left(xy+yz+zx\right)\ge0\)

hay: \(x^2+y^2+z^2\ge xy+yz+zx\)

Vậy để đẳng thức (1) xảy ra thì x = y = z = 2.

2/ (x-z)^2 - 1 - (y^2-2y)

= (x-z+1)(x-z-1)-y(y-2) 

(a + b + c)²

= (a +b +c)(a + b+c)

= a² + ab + ac + ba +b² + bc + ca + cb +c²

= a² + b² + c² + 2ab + 2bc + 2ac

(3x + 2)² - (x + 2)(9x - 1)

= (3x)² + 2.3x.2 + 2² - (9x² - x + 18x - 2)

= 9x² + 12x + 4 - 9x² + x - 18x + 2

= - 5x + 6

\(\left(3x+2\right)^2-\left(x+2\right)\left(9x-1\right)\)

\(=\left(9x^2+6x+4\right)-\left(9x^2-x+18x-2\right)\)

\(=9x^2+6x+4-9x^2-17x+2\)

\(=6-11x\)

\(2A=2x^2+2y^2-2xy+2x+2y\)

\(2A=x^2-2xy+y^2+x^2+2x+1-1+y^2+2y+1-1\)

\(2A=\left(x^2-2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)-1-1\)

\(2A=\left(x-y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2-2\ge-2\)

\(\Rightarrow2A\le-2\Rightarrow A\le-1\)

\(\Rightarrow A_{min}=-1\Leftrightarrow\hept{\begin{cases}x-y=0\\x+1=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\x=-1\\y=-1\end{cases}\Rightarrow}x=y=-1}\)

CHÚC BẠN HỌC TỐT NHÉ 

nha ! CẢM ƠN!!!!!

ủa vậy nó =0 rồi bạn ơi

x^3 +y^3 +z^3 -x^3 -y^3 -z^3 =0 rồi 

cần xem lại đề nha

thấy mình nói đúng thi T I C K cho mình nha mấy bạn 

(x^3 +y^3 + z^3) - (x^3 +y^3 +z^3)

= (1-1) (x^3 + y^3 + z^3)

= 0 *(x^3 + y^3 + z^3)

Sao kì quá =.=!!!

(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)c(a+b+c)(a+b+c)3=((a+b)+c)3=(a+b)3+c3+3(a+b)c(a+b+c)
=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)=a3+b3+3ab(a+b)+c3+3(a+b)c(a+b+c)
=a3+b3+c3+3(a+b)(ab+c(a+b+c))=a3+b3+c3+3(a+b)(ab+c(a+b+c))
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)(a+c)(b+c)=a3+b3+c3+3(a+b)(a+c)(b+c)

~~

(a+b+c)³=(a+b)³+3(a+b)²c+3(a+b)c²+c³

=a³+b³+c³+3a²b+3ab²+3(a²+2ab+b²)c+3ac²+3bc²

=a³+b³+c³+3a²b+3ab²+3a²c+6abc+3b²c+3ac²+3bc²

(a+b)(b+c)(a+c)=(a+b)(b+c)a+(a+b)(b+c)c

=(a+b)ab+(a+b)ac+(a+b)bc+(a+b)c²

=a²b+ab²+a²c+abc+abc+b²c+ac²+bc²

=>3(a+b)(b+c)(c+a)=3a²b+3ab²+6abc+3b²c+3ac²+3bc²

=>(a+b+c)³=a³+b³+c³+3(a+b)(b+c)(c+a)

=>đpcm