toán 8 khó đây

L

linhhuyenvuong

\frac{1}{a}+ \frac{1}{b}+ \frac{1}{c}=0
rút gọn
\frac{1}{a^2+2bc}+\frac{1}+{b^2+2ac}+\frac{1}{c^2+2ab}

_____________________________________
TA có
[tex]\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0[/tex]
=> ab+bc+ac=0
a^2+2bc=a^2+bc+(-ab-ac)=(a-b)(a-c)
tương tự b^2+2ac=(b-a)(b-c)
c^2+2ab=(c-a)(c-b)
=> A=[tex]\frac{1}{(a-b)(a-c)}+\frac{1}{(b-a)(b-c)}+\frac{1}{(c-a)(c-b)}[/tex]
=[tex]\frac{b-c+c-a+a-b}{(a-b)(b-c)(a-c)}[/tex]
=0
 
Last edited by a moderator:
H

hell_angel_1997

nhờ mod xoá giùm bài này
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
 
Last edited by a moderator:
Top Bottom