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...

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

Tương tự \(x^2+z^2-y^2=-2xz;y^2+z^2-x^2=-2yz\)

Cộng VTV:

\(\Leftrightarrow\text{Biểu thức }=\dfrac{xy}{-2xy}+\dfrac{xz}{-2xz}+\dfrac{yz}{-2yz}=-\dfrac{1}{8}\)

x+y+z=0

=>x+y=-z; x+z=-y; y+z=-x

\(x^2+y^2-z^2\)

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

\(=\left(-z\right)^2-2xy-z^2=-2xy\)

\(x^2+z^2-y^2\)

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

\(=\left(-y\right)^2-2xz-y^2=-2xz\)

\(y^2+z^2-x^2\)

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

\(=\left(-x\right)^2-2yz-x^2=-2yz\)

\(\frac{xy}{x^2+y^2-z^2}+\frac{xz}{x^2+z^2-y^2}+\frac{yz}{y^2+z^2-x^2}\)

\(=\frac{xy}{-2xy}+\frac{xz}{-2xz}+\frac{yz}{-2yz}\)

\(=-\frac12-\frac12-\frac12=-\frac32\)

 Trước hết, ta đi chứng minh một bổ đề sau: Nếu \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\). Thật vậy, ta phân tích 

 \(P=a^3+b^3+c^3-3abc\)

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

\(P=\left(a+b+c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)

\(P=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).

Hiển nhiên nếu \(a+b+c=0\) thì \(P=0\) hay \(a^3+b^3+c^3=3abc\), bổ đề được chứng minh.

Do \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\) nên áp dụng bổ đề, ta được \(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\).

Vì vậy \(\dfrac{yz}{x^2}+\dfrac{zx}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\) \(=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)\) \(=xyz.\dfrac{3}{xyz}=3\). Ta có đpcm

Bài này ez thôi, làm mãi rồi.

Theo đề bài, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

=>\(\dfrac{xy+yz+xz}{xyz}=0\)

=> xy+yz+zx=0

=> \(\left\{{}\begin{matrix}xy=-yz-zx\\yz=-xy-zx\\zx=-xy-yz\end{matrix}\right.\)

Ta có: x²+2yz=x²+yz-xy-zx=(x-y)(x-z)

           y²+2xz=y²+xz-xy-yz=(x-y)(z-y)

           z²+2xy=z²+xy-yz-xz=(x-z)(y-z)

=> \(\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(x-y\right)\left(z-y\right)}+\dfrac{xy}{\left(x-z\right)\left(y-z\right)}=\dfrac{yz\left(y-z\right)-xz\left(x-z\right)+xy\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=\dfrac{\left(x-y\right)\left(x-z\right)\left(y-z\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}=1\)

Cảm ơn, cậu giỏi quá!!! Thông cảm cho đứa ngu toán

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

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow\dfrac{xy+yz+xz}{xyz}=0\Leftrightarrow xy+yz+xz=0\Leftrightarrow yz=-xy-xz\)

Ta có \(x^2+2yz=x^2+yz-xy-xz=\left(x-y\right)\left(x-z\right)\)

Tương tự \(y^2+2xz=\left(y-x\right)\left(y-z\right);z^2-2xy=\left(z-x\right)\left(z-y\right)\)

\(A=\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xz}{\left(y-z\right)\left(y-x\right)}+\dfrac{xy}{\left(z-x\right)\left(z-y\right)}\\ A=\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{-yz\left(y-z\right)+xz\left(y-z\right)+xz\left(x-y\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(y-z\right)\left(xz-yz\right)+\left(x-y\right)\left(xz-xy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\\ A=\dfrac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)

⇔xy+yz+zx=0

=yz/(x−y)(x−z)

Tương tự: 

; 

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)

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

Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)

\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)