Toán Toán 7: Tổng hợp

T

thangvegeta1604

a+b+c=0\Rightarrow $(a+b+c)^2=0$
\Rightarrow $a^2+b^2+c^2+2(ab+bc+ca)=0$
\Rightarrow $2(ab+bc+ca)=-(a^2+b^2+c^2)$.
Mà $a^2+b^2+c^2$\geq 0\Rightarrow $-(a^2+b^2+c^2)$\leq 0.
Do đó: $2(ab+bc+ca)$\leq 0
\Rightarrow $ab+bc+ca$\leq 0.
 
L

lethuoc

a+b+c=0, chung minh ac+bc+ca> hoac= 0

a+b+c=0,vay a=-(b+c)
ab+bc+ca=-(b+c)b+bc+-(b+c)c=-b^2-bc+bc-bc-c^2
=-b^2-bc-c^2=(-b^2-b^2-2bc-c^2-c^2)/2
=-(b^2+c^2)/2-[(b+c)^2]/2 luon luon > hoac = 0
 
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