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![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 : \(M=\frac{1}{a+b+c},a+b+c\ne0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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:
\(A=\dfrac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(B=bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2\)
\(=bcy^2+bcz^2+caz^2+cax^2+aby^2-2\left(bcyz+acxz+abxy\right)\) (1)
Từ giả thiết suy ra:
\(a^2x^2+b^2y^2+c^2z^2+2\left(bcyz+acxz+abxy\right)=0\) (2)
Từ (1) và (2) suy ra:
\(B=ax^2\left(b+c\right)+by^2\left(a+c\right)+cz^2\left(a+b\right)+a^2x^2+b^2y^2+c^2z^2\)
\(=ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)\)
Do đó: \(A=\dfrac{B}{ax^2+by^2+cz^2}=a+b+c\)
Đặt: B = bc(y-z)2 + ca(z-x)2 + ab(x-y)2
= bcy2 + bcz2 + caz2 + cax2 + abx2 + aby2 - 2(bcyz + acxz + abxy) (1)
=> a2x2 + b2y2 + c2z2 + 2(bcyz + acxz + abxy) = 0 (2)
Từ (1) và (2) suy ra:
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)
= (az2+by2+cz2)(a+b+c)
Vậy \(A=\dfrac{B}{ax^2+by^2+cz^2}=a+b+c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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