[Toán 8]Chứng minh đẳng thức

K

khanhtoan_qb

1. Cho ax + by + cz = 0
c/m : [TEX]\frac{a{x}^{2}+b{y}^{2}+c{z}^{2}}{bc{(y-z)}^{2}+ca{(z-x)}^{2}+ab{(x-y)}^{2}}=\frac{1}{a+b+c}[/TEX]
Ta có
[TEX]ax + by + cz = 0 \Rightarrow a^2x^2 + b^2y^2 + c^2z^2 + 2abxy + 2bcyz + 2 acxz = 0[/TEX]
\Rightarrow[TEX]a^2x^2 + b^2y^2 + c^2z^2 = -2 (abxy + bcyz + acxz) [/TEX]
[TEX]\frac{a{x}^{2}+b{y}^{2}+c{z}^{2}}{bc{(y-z)}^{2}+ca{(z-x)}^{2}+ab{(x-y)}^{2}}[/TEX]
xét mẫu [TEX]A = bc(y-z)^2+ca(z-x)^2+ab(x-y)^2 [/TEX]
[TEX] = bc (y^2 - 2xy + z^2) + ca(z^2 - 2xz + x^2) + ab (x^2 - 2xy + y^2)[/TEX]
[TEX] = bcy^2 + bcz^2 + caz^2 + cax^2 + abx^2 + aby^2 -2 (abxy + bcyz + acxz) [/TEX]
[TEX] = bcy^2 + bcz^2 + caz^2 + cax^2 + abx^2 + aby^2 + a^2x^2 + b^2y^2 + c^2z^2 [/TEX]
[TEX] = a^2x^2 + abx^2 + acx^2 + aby^2 + b^2y^2 + bcy^2 + acz^2 + bcz^2 + c^2z^2[/TEX]
[TEX]= (ax^2 + by^2 + cz^2 )(a + b + c)[/TEX]
Từ đó
\Rightarrow đpcm:):)
 
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