TOan kho day,ạ gjoj thj vao gjaj nha!!!

K

khanh_ndd

1)CHo a+b+c=1;a^2+b^2+c^2=1;x/a=y/b=z/c=k
Tinh xy+yz+zx
Từ gt dùng t/c tỉ lệ thức [TEX]\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=k\Rightarrow x+y+z=k[/TEX] và
[TEX]\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}= \frac{x^2+y^2+z^2}{a^2+b^2+c^2}=k^2 \Rightarrow x^2+y^2+z^2=k^2[/TEX]
[TEX]\Rightarrow xy+yz+zx=\frac{(x+y+z)^2-(x^2+y^2+z^2)}{2}=0[/TEX]
 
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M

mini_baby_237

ko hieu

Toj ko hju:xy+yz+zx=(a+b+c)^2-(a^2+b^2+c^2)/2.
Toj co cach gjaj #:
Dat x/a=y/b=z/c=k
x=ak,y=bk;z=ck
laj co:ab+bc+ac=(a+b+c)^2-(a^2+b^2+c^2)/2=0
suy ra xy+yz+zx=0
 
S

soulma90

minh lam` ne`
(a+b+c)2 = a2 +b2+c2 +2ab+2ac+2bc
1 = 1+2ab+2ac+2bc
2ab+2ac+2bc =0
ab+ac+bc = 0
 
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