Cho ax+by+cz=0 và a+b+c =1/2018 Chứng minh rằng \(\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}\) =2018
Cho ax+by+cz=0 và a+b+c =1/2018 Chứng minh : \(\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}=2018\)
\(Cho\) \(ax+by+cz=0;a+b+c=\dfrac{1}{2018}\) . CMR: \(\dfrac{ax^{2\:}+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}=2018\)
cho ax+by+cz=0,a+b+c=2015. tính Q=\(\frac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2}\)
Cho ax+by+cz=0; a+b+c =\(\dfrac{2019}{2018}\)
Tính : \(P=\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}\)
Bạn tham khảo bài tương tự tại đây:
Giả sử : \(ax+by+cz=0.\)
Chứng minh : \(\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2}=\dfrac{1}{a+b+c}\)
\(ax+by+cz=0\Rightarrow\left(ax+by+cz\right)^2=0\)
\(\Rightarrow a^2x^2+b^2y^2+c^2z^2=-2\left(axby+bycz+axcz\right)\)
Ta co
\(\dfrac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(z-x\right)^2+ab\left(x-y\right)^2}\)
\(=\dfrac{ax^2+by^2+cz^2}{bcy^2-2bcyz+bcz^2+acz^2-2aczx+acx^2+abx^2-2abxy+aby^2}\)
\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2-2\left(axby+bcyz+axcz\right)}\)
\(=\dfrac{ax^2+by^2+cz^2}{bcy^2+bcz^2+acz^2+acx^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2}\)
\(=\dfrac{ax^2+by^2+cz^2}{\left(acx^2+abx^2+a^2x^2\right)+\left(bcy^2+aby^2+b^2y^2\right)+\left(c^2z^2+acz^2+bcz^2\right)}\)
\(=\dfrac{ax^2+by^2+cz^2}{ax^2\left(a+b+c\right)+by^2\left(a+b+c\right)+cz^2\left(a+b+c\right)}\)
\(=\dfrac{ax^2+by^2+cz^2}{\left(a+b+c\right)\left(ax^2+by^2+cz^2\right)}=\dfrac{1}{a+b+c}\) ( dpcm)
Cho biết: ax+by+cz=0. Rút gọn: \(A=\dfrac{bc.\left(y-z\right)^2+ca.\left(z-x\right)^2+ab.\left(x-y\right)^2}{ax^2+by^2+cz^2}\)
\(A=\dfrac{bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2-2bcyz-2cazx-2abxy}{ax^2+by^2+cz^2}=\dfrac{\left(bcy^2+bcz^2+caz^2+cax^2+abx^2+aby^2+a^2x^2+b^2y^2+c^2z^2\right)-\left(ax+by+cz\right)^2}{ax^2+by^2+cz^2}=\dfrac{\left(ax^2+by^2+cz^2\right)\left(a+b+c\right)}{ax^2+by^2+cz^2}=a+b+c\)
Cho biết ax + by + cz = 0
Rút gọn: \(A=\frac{bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2}{ax^2+by^2+cz^2}\)
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:
\(A=\frac{B}{ax^2+by^2+cz^2}=a+b+c\)
Cho \(\left\{{}\begin{matrix}ax+by+cz=0\\a+b+c=\frac{1}{2019}\end{matrix}\right.\) . Tính giá trị của \(\frac{ax^2+by^2+cz^2}{ab\left(x-y\right)^2+bc\left(y-z\right)^2+ca\left(z-x\right)^2}\)
Biết ax+by+cz=0. Rút gọn: \(\frac{ax^2+by^2+cz^2}{bc\left(y-z\right)^2+ca\left(z-x\right)^2+ab\left(x-y\right)^2}\)
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