Đặt B là mẫu thức của P thì :

B = ab(x - y)² + bc(y - z)² + ca(z - x)² = abx² - 2abxy + aby² + bcy² - 2bcyz + bcz² + caz² - 2cazx + cax²

   = ax²(b + c) + by²(a + c) + cz²(a + b) - 2(bcyz + acxz + abxy) (1)

ax + by + cz = 0 => (ax + by + cz)² = 0 <=> a²x² + b²y² + c²z² + 2(bcyz + acxz + abxy) = 0 

=> -2(bcyz + acxz + abxy) = a²x² + b²y² + c²z² (2)

Từ (1) và (2),ta có : B = ax²(b + c) + by²(a + c) + cz²(a + b) + a²x² + b²y² + c²z²

= ax²(a + b + c) + by²(a + b + c) + cz²(a + b + c) = (a + b + c)(ax² + by² + cz²)

\(\Rightarrow P=\frac{1}{a+b+c}=2017\)

P=2017

Khó thế

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)

Giải

Ta có: \(B=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)

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

\(=ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)-2\left(bcyz+acxz+abxy\right)\)(1)

Từ giả thiết suy ra:

\(a^2x^2+b^2y^2+c^2z^2+2\left(abxy+acxz+bcyz\right)=0\) (2)

Từ (1) và (2):

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

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

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

Do đó:

\(A=\frac{B}{ax^2+by^2+cz^2}=a+b+c\)

Giải:
Chú ý sử dụng hằng đẳng thức : 
(m+n+p)² = m² + n² + p² + 2mn + 2mp + 2np
Áp dụng hằng đẳng thức trên, ta có:
ax+by+cz = 0 ⇒ (ax+by+cz)² = 0

⇒a²x²+b²y²+c²z²+2ax.by+2ax.cz+2by.cz=0

⇒a²x²+b²y²+c²z²= − (2abxy+2aczx+2bcyz)

Ta lại có:
bc.(y−z)²+ac.(x−z)²+ab.(x−y)²

=bc(y²−2yz+z²)+ac(x²−2xz+z²)+ab(x²−2xy+y²)

=bcy²+bcz²−2bcyz+acx²+acz²−2acxz+abx²+aby²−2abxy

=(bcy²+bcz²+acx²+acz²+abx²+aby²)−(2abxy+2aczx+2bcyz)

=bcy²+bcz²+acx²+acz²+abx²+aby²+a²x2+b²y²+c²z²

=x²(ac+ab+a²)+y²(bc+ab+b²)+z²(bc+ac+c²)

=ax²(a+b+c)+by²(a+b+c)+cz²(a+b+c)

=(a+b+c)(a.x²+b.y²+c.z²)

Vậy:
A=a.x²+b.y²+c.z2bc.(y−z)²+ac.(x−z)²+ab.(x−y)²=ax²+by²+cz²(a+b+c)(a.x²+b.y²+c.z²)

A=1a+b+cA=1a+b+c