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

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:

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

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2axcz=0\)

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

Ta có:

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

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

\(=bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2axby-2bycz-2axcz\)

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

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

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

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

Thay vào A ta được:

\(A=\dfrac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)

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

Lương Ngọc Anh nhanh nhỉ 

Ta có : \(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\Rightarrow a^2x^2+b^2y^2+c^2z^2+2axby+2bycz+2czax=0\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2abxy-2bycz-2czax\)

Xét tử số :  \(bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2=bc\left(y^2-2yz+z^2\right)+ac\left(z^2-2xz+x^2\right)+ab\left(x^2-2xy+y^2\right)\)\(=bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(bcyz+abxy+acxz\right)\)

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

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

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

\(\Rightarrow A=\frac{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}=\frac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)

x^20+(x+1)^11=2016^y=?

Từ giả thiết ta có: \(ax+by+cz=0\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)

Ta biến đổi mẫu của biểu thức A: 

\(bc\left(y^2-2yz+z^2\right)+ac\left(x^2-2xz+z^2\right)+ab\left(x^2-2xy+y^2\right)\)

\(=bcy^2+bcz^2+acx^2+acz^2+abx^2+aby^2-2\left(bycz+axcz+axby\right)\)

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

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

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

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

Vậy  \(A=\frac{ax^2+by^2+cz^2}{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}=\frac{1}{a+b+c}\)

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:

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

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

\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)\)( 1 )

Mà  \(a.x+by+cz=0\)

\(\Rightarrow\left(a.x+by+cz\right)^2=0^2\)

\(\Rightarrow a^2x^2+b^2y^2+c^2z^2+2\left(axby+axcz+bycz\right)=0\)( 2 )

\(\left(1\right)\left(2\right)\Rightarrow B=B+0\)

\(=bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2\left(bcyz+acxz+abxy\right)+a^2x^2+b^2y^2+c^2z^2+2\left(axby+axcz+bycz\right)\)

\(=a.x^2\left(b+c\right)+b.y^2\left(a+c\right)+c.z^2\left(a+b\right)+a^2x^2+b^2y^2+z^2c^2\)

\(=a.x^2\left(a+b+c\right)+b.y^2\left(a+b+c\right)+cz^2\left(a+b+c\right)\)

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

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

Vậy ...

x^20+(x+1)^11=2016^y=?

\(\frac{1}{a+b+c}\)