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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:
\(A=\frac{B}{ax^2+by^2+cz^2}=a+b+c\)
theo đề bài: \(ax+by+cz=0\)=> \(\left(ax+by+cz\right)^2=0\)
=> \(a^2x^2+b^2y^2+c^2z^2+2\left(axby+bycz+axcz\right)=0\left(1\right)\)
ta lại có tử số =\(bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)
=\(bcy^2+bcz^2+caz^2+acx^2+abx^2+aby^2-2\left(abxy+acxz+bcyz\right)\)(2)
từ (1)(2)=>
Tử số=\(ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)+a^2x^2+b^2y^2+c^2z^2\)
=\(\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)\)
vậy A=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
Ta có: \(ax+by+cz=0\)
\(\Rightarrow\left(ax+by+cz\right)^2=0\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2\left(axby+bycz+axcz\right)=0\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)
Lại có: \(\frac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}\)
\(=\frac{ax^2+by^2+cz^2}{bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)}\)
\(=\frac{ax^2+by^2+cz^2}{bcy^2-2bcyz+bcz^2+acx^2-2acxz+acz^2+abx^2-2abxy+aby^2}\)
\(=\frac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)}\)
\(=\frac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2}\)
\(=\frac{ax^2+by^2+cz^2}{\left(acx^2+abx^2+a^2x^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(bcz^2+acz^2+c^2z^2\right)}\)
\(=\frac{ax^2+by^2+cz^2}{ax^2\left(c+b+a\right)+by^2\left(c+a+b\right)+cz^2\left(b+a+c\right)}\)
\(=\frac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\frac{1}{a+b+c}=\frac{1}{\frac{1}{2020}}=2020\) (đpcm)