Lời giải:

Từ \(ax+by+cz=0\Rightarrow (ax+by+cz)^2=0\)

\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2=-2(axby+axcz+bycz)\)

\(=-2(bcyz+cazx+abxy)\)

Khi đó:

\(bc(y-z)^2+ca(z-x)^2+ab(x-y)^2=bc(y^2-2yz+z^2)+ca(z^2-2zx+x^2)+ab(x^2-2xy+y^2)\)

\(=(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2)-(2bcyz+2cazx+2abxy)\)

\(=(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2)+(a^2x^2+b^2y^2+c^2z^2)\)

\(=ax^2(a+b+c)+by^2(a+b+c)+cz^2(a+b+c)=(a+b+c)(ax^2+by^2+cz^2)\)

Do đó:

\(\frac{ax^2+by^2+cz^2}{bc(y-z)^2+ca(z-x)^2+ab(x-y)^2}=\frac{ax^2+by^2+cz^2}{(ax^2+by^2+c^2)(a+b+c)}=\frac{1}{a+b+c}=\frac{1}{\frac{1}{2019}}=2019\)

Ta có đpcm.

\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)

Đặt \(A=\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+cz\left(z-x\right)}\)

Từ ax+by+cz=0

=>(ax+by+cz)²=0

=>a²x²+b²y²+c²z²+2axby+2bycz+2czax=0

=>a²x²+b²y²+c²z²=-2(ax+by+byca+czax)

Xét mẫu thức: \(ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2\)

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

\(=abx^2-2abxy+aby^2+bcy^2-2bcyz+bcz^2+caz^2-2cazx+cax^2\)

\(=\left(abx^2+bcz^2\right)+\left(aby^2+acz^2\right)+\left(acx^2+bcy^2\right)-2\left(abxy+bcyz+cazx\right)\)

\(=\left(aby^2+acz^2\right)+\left(abx^2+bcz^2\right)+\left(acx^2+bcy^2\right)+a^2x^2+b^2y^2+c^2z^2\)

\(=\left(a^2x^2+aby^2+acz^2\right)+\left(abx^2+b^2y^2+bcz^2\right)+\left(acx^2+bcy^2+c^2z^2\right)\)

\(=a\left(ax^2+by^2+cz^2\right)+b\left(ax^2+by^2+cz^2\right)+c\left(ax^2+by^2+cz^2\right)\)

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)

Do đó: \(A=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{\frac{1}{2018}}=2018\) (dpcm)

Phân tích mẫu :

\(M=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

Khai triển các bình phương và gom các nhân tử chung :

\(M=\left(ab+ac\right)x^2+\left(ab+bc\right)y^2+\left(bc+ac\right)z^2-2abxy-2bcxy-2acxy\)

\(=\left[\left(ab+ac\right)x^2+a^2x^2+\left(ab+bc\right)y^2+b^2y^2+\left(bc+ac\right)z^2+c^2z^2\right]-\)\(\left(a^2x^2+b^2y^2+c^2z^2+2ab+2aczx+2bcyz\right)\)

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)-\left(ax+by+cz\right)^2\)

\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\) ( vì \(ax+by+cz=0\) )

Kết quả :  \(M=\frac{1}{a+b+c},a+b+c\ne0\)