giup tui voi ban oi

M

ms.sun

thocon2011 said:
Cho [tex] a+b+c=1[/tex]
[tex]a^2+b^2+c^2=1[/tex]
[tex] \frac{x}{a} = \frac{y}{b}= \frac{z}{c}[/tex]

Chứng minh rằng: xy+yz+xz=0:p:p:p:p:p:p:p:p
Chú ý latex
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[TEX]a+b+c=1 \Rightarrow (a+b+c)^2=1 [/TEX]
mà [TEX]a^2+b^2+c^2=1 \Rightarrow ab+bc+ca=0[/TEX]
vì [TEX] \frac{x}{a}=\frac{y}{b}=\frac{z}{c} \Rightarrow \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c} =x+y+z[/TEX]
[TEX] \Rightarrow x=a(x+y+z) ; y=b(x+y+z); z=c(x+y+z)[/TEX]
vậy [TEX]xy+yz+zx = a(x+y+z).b(x+y+z) +b(x+y+z).c(x+y+z)+c(x+y+z).a(x+y+z)[/TEX]
[TEX] \Leftrightarrow xy+yz+zx = ab (x+y+z)^2+bc(x+y+z)^2+ca(x+y+z)^2[/TEX]
[TEX] \Leftrightarrow xy+yz+zx=(x+y+z)^2(ab+bc+ca) =0 (dpcm)[/TEX]

ok chưa ;))
 
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