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Câu hỏi của Rồng Con - Toán lớp 8 | Học trực tuyến
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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)
![](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)
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)
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:
\(A=\dfrac{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}{ax^2+by^2+cz^2}=a+b+c\)