[toán 8 đại số] Tính giá trị biểu thức.

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pipilove_khanh_huyen

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vipboycodon

$N = \dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}$

$N =\dfrac{x}{x+ax}+\dfrac{y}{y+by}+\dfrac{z}{z+cz}$

$N =\dfrac{by+cz}{ax+by+cz}+\dfrac{ax+cz}{ax+by+cz}+ \dfrac{ax+by}{ax+by+cz}$

$N =\dfrac{2(ax+by+cz)}{ax+by+cz}$

$N =\dfrac{by+cz+ax+cz+ax+by}{ax+by+cz}$

$N = 2$
 
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phuong_july

Cho x= by+cz
y= ax+cz
z= ax+by
và x+y+z khác 0
Tính N=1/a+1 + 1/1+b +1/1+c
thanks liền cho ai giải đc
Ta có [TEX]x+y+z=by+cz+ax+cz+ax+by= 2(ax + by + cz) [/TEX]
Thay: [TEX]z=ax+by \Rightarrow x+y+z= 2(z+cz)= 2z(1+c) [/TEX]
[TEX]\Rightarrow \frac{1}{1+c}=\frac{2z}{x+y+z}[/TEX]

Tương tự ta có: [TEX]\frac{1}{1+a}=\frac{2x}{x+y+z}[/TEX]
[TEX]\frac{1}{1+b}=\frac{2y}{x+y+z}[/TEX]

[TEX]\Rightarrow N= \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=\frac{2(x+y+z)}{x+y+z}=2[/TEX]
 
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