Ta có : \(x^2-xy=y^2-yz=z^2-zx\)Cộng 3 vế , suy ra :

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

Do \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}< =>\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0}\)

Dấu = xảy ra khi và chỉ khi \(\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}< =>x=y=z}\)

Khi đó ta được : \(M=\frac{x}{z}+\frac{z}{y}+\frac{y}{x}=1+1+1=3\)( do x=y=z )

Bạn ơi đề bài cho a khác 0 mà bạn

ta có : xy + yz +zx = 0

        * yz = -xy-zx

\(\Rightarrow\)*xy = - yz - zx

         *zx= -xy-yz

ta có : M = \(\frac{xy}{z}+\frac{zx}{y}+\frac{yz}{x}\)

          M = \(\frac{-yz-zx}{z}+\frac{-xy-yz}{y}+\frac{-xy-zx}{x}\)

          M = \(\frac{z\times\left(-y-x\right)}{z}+\frac{y\times\left(-x-z\right)}{y}+\frac{x\times\left(-y-z\right)}{x}\)

          M = -y - x - x - z - y - z

         M = -2y - 2x - 2z

         M = -2( x+y+z )

   mà x+y+z=-1

         M = (-2) . (-1)

         M =2

 Quản lý

Đặt \(\frac{x^2 - yz}{a} = \frac{y^2 - zx}{b} = \frac{z^2 - xy}{c} = k\) (k<>0)

=>\(a=\frac{x^2 - yz}{k},\quad b=\frac{y^2 - zx}{k},\quad c=\frac{z^2-xy}{k}\)

\(a^2 = \frac{(x^2 - yz)^2}{k^2} = \frac{x^4 - 2x^2yz + y^2z^2}{k^2}\)

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

\(a^2 - bc = \frac{(x^4 - 2x^2yz + y^2z^2) - (y^2z^2 - xy^3 - xz^3 + x^2yz)}{k^2}\)

\(=\frac{x^4 - 3x^2yz + xy^3 + xz^3}{k^2}\)

\(=\frac{x(x^3 + y^3 + z^3 - 3xyz)}{k^2}\)

=>\(\frac{a^2 - bc}{x} = \frac{x^3 + y^3 + z^3 - 3xyz}{k^2} \quad (1)\)

\(b^2 = \frac{(y^2 - zx)^2}{k^2} = \frac{y^4 - 2xy^2z + x^2z^2}{k^2}\)

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

=>\(b^2 - ca = \frac{(y^4 - 2xy^2z + x^2z^2) - (x^2z^2 - yz^3 - x^3y + xy^2z)}{k^2}\)

\(=\frac{y^4 - 3xy^2z + yz^3 + x^3y}{k^2}\)

\(=\frac{y(x^3 + y^3 + z^3 - 3xyz)}{k^2}\)

=>\(\frac{b^2 - ca}{y} = \frac{x^3 + y^3 + z^3 - 3xyz}{k^2} \quad (2)\)

\(c^2 = \frac{(z^2 - xy)^2}{k^2} = \frac{z^4 - 2xyz^2 + x^2y^2}{k^2}\)

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

=>\(c^2 - ab = \frac{(z^4 - 2xyz^2 + x^2y^2) - (x^2y^2 - x^3z - y^3z + xyz^2)}{k^2}\)

\(=\frac{z^4 - 3xyz^2 + x^3z + y^3z}{k^2}\)

\(=\frac{z(x^3 + y^3 + z^3 - 3xyz)}{k^2}\)

=>\(\frac{c^2 - ab}{z} = \frac{x^3 + y^3 + z^3 - 3xyz}{k^2} \quad (3)\)

Từ (1),(2),(3) suy ra \(\frac{a^2 - bc}{x} = \frac{b^2 - ca}{y} = \frac{c^2 - ab}{z} = \frac{x^3 + y^3 + z^3 - 3xyz}{k^2}\)

Đặt \(\frac{x^2 - yz}{a} = \frac{y^2 - zx}{b} = \frac{z^2 - xy}{c} = k\) (k<>0)

=>\(a=\frac{x^2 - yz}{k},\quad b=\frac{y^2 - zx}{k},\quad c=\frac{z^2-xy}{k}\)

\(a^2-bc=0\)

=>\(\left(\frac{x^2 - yz}{k}\right)^2 = \left(\frac{y^2 - zx}{k}\right) \cdot \left(\frac{z^2 - xy}{k}\right)\)

=>\((x^2 - yz)^2 = (y^2 - zx)(z^2 - xy)\)

=>\(x^4 - 2x^2yz + y^2z^2 = y^2z^2 - xy^3 - xz^3 + x^2yz\)

=>\(x^4 - 2x^2yz = -xy^3 - xz^3 + x^2yz\)

=>\(x^4 - 3x^2yz + xy^3 + xz^3 = 0\)

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

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

mà x+y+z=2010

nên \(x^2+y^2+z^2-xy-xz-yz=0\)

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

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

=>x=y=z

mà x+y+z=2010

nên x=y=z=2010/3=670

Ta có \(\dfrac{\left(x^2-yz\right)^2}{a^2}=\dfrac{\left(y^2-zx\right)\left(z^2-xy\right)}{bc}\) mà a² = bc nên:

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

\(\Leftrightarrow x^4+y^2z^2-2x^2yz=y^2z^2+x^2yz-xy^3-xz^3\)

\(\Leftrightarrow x^4+xy^3+xz^3-3x^2yz=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x^3+y^3+z^3=3xyz\end{matrix}\right.\).

Rõ ràng nếu \(x^3+y^3+z^3=3xyz\) thì \(x=y=z\) (tính chất quen thuộc). Do đó \(\dfrac{x^2-yz}{a}=0\) (vô lí).

Do đó x = 0.

Kết hợp với x + y + z = 2010 thì y + z = 2010.

Rõ ràng với mọi x, y, z thỏa mãn y + z = 2010 và x = 0 thì ta thấy thỏa mãn đk bài toán.

Vậy...