\(x+y+z=0\)

\(x+y=-z\)

\(\left(x+y\right)^3=-z^3\)

\(x^3+3xy\left(x+y\right)+y^3=-z^3\)

\(x^3+\left(-3xyz\right)+y^3=-z^3\)

\(x^3+y^3+z^3=3xyz\)( đpcm )

x+y+z = 0

<=> x+y = -z

<=> (x+y)^3 = -z^3

<=> x^3+y^3+3xy.(x+y) = -z^3

<=> x^3+y^3+z^3 = -3xy.(x+y)

Mà x+y+z = 0 => x+y = -z

=> x^3+y^3+z^3 = -3xy.(-z) = 3xyz

=> ĐPCM

k mk nha

Bạn chứng minh hằng đẳng thức này là sẽ làm được

\(x^3+y^3+z^3-3xyz=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

Mà \(x+y+z=0\)

\(\Rightarrow x^3+y^3+z^3-3xyz=0\)

\(\Rightarrow x^3+y^3+z^3=3xyz\)

Bài 2:

a, \(x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(x+y\right)z-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)^3-3\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xy-3yz-3zx\right]\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)

2a ) Ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)

Ta có: \(\frac{x^3+y^3+z^3-3xyz}{x+y+z}\)

\(=\frac{\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz}{x+y+z}\)

\(=\frac{\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)}{x+y+z}\)

\(=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-zx-3xy\right)}{x+y+z}\)

\(=x^2+y^2+z^2-xy-yz-zx=\frac{1}{2}\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]\ge0\left(\forall x,y,z\right)\)

=> đpcm

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mn giúp mình với

Sửa đề: Chứng minh x=y=z

\(x^3+y^3+z^3=3xyz\)

=>\(\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz=0\)

=>\(\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

=>\(\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-xz-yz-3xy\right)=0\)

=>\(x^2+y^2+z^2-xy-xz-yz=0\)

=>\(2x^2+2y^2+2z^2-2xy-2xz-2yz=0\)

=>\(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2=0\)

=>\(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

=>x=y=z

Có:

\(x^3+y^3+z^3=3xyz\\\Leftrightarrow x^3+y^3+z^3-3xyz=0\\\Leftrightarrow(x+y)^3+z^3-3xy(x+y)-3xyz=0\\\Leftrightarrow (x+y+z)^3-3(x+y)z(x+y+z)-3xy(x+y+z)=0\\\Leftrightarrow (x+y+z)[(x+y+z)^2-3(x+y)z-3xy]=0\\\Leftrightarrow (x+y+z)(x^2+y^2+z^2+2xy+2yz+2xz-3xz-3yz-3xy)=0\\\Leftrightarrow (x+y+z)(x^2+y^2+z^2-xy-yz-xz)=0\\\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0 (vì.x+y+z\neq0)\\\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2xz=0\\\Leftrightarrow(x^2-2xy+y^2)+(y^2-2yz+z^2)+(x^2-2xz+z^2)=0\\\Leftrightarrow(x-y)^2+(y-z)^2+(x-z)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(x-y\right)^2\ge0\forall x;y\\\left(y-z\right)^2\ge0\forall x;y\\\left(x-z\right)^2\ge0\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2\ge0\forall x;y;z\)

Mà: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

nên: \(\left\{{}\begin{matrix}x-y=0\\y-z=0\\x-z=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=y\\y=z\\x=z\end{matrix}\right.\Leftrightarrow x=y=z\left(đpcm\right)\)

\(Toru\)

2

a

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^3=\left(-z\right)^3\)

\(\Rightarrow x^3+y^3+3x^2y+3xy^2=-z^3\)

\(\Rightarrow x^3+y^3+z^3=3xy\left(x+y\right)=3xyz\)

b

Đặt \(a-b=x;b-c=y;c-a=z\Rightarrow x+y+z=0\)

Ta có bài toán mới:Cho \(x+y+z=0\).Phân tích đa thức thành nhân tử:\(x^3+y^3+z^3\)

Áp dụng kết quả câu a ta được:

\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)