{Toán 8}Toán nâng cao

Q

quocdat25

A

anhbez9

chém câu dễ:câu 2:
ta có:abc=1
\Rightarrow[TEX]N=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{c}{ac+c+abc}=\frac{a}{a(bc+b+1)}+\frac{b}{bc+b+1}+\frac{c}{c(a+ab+abc)}=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{a(bc+b+1)}=\frac{a+ab+abc}{a(bc+b+1)}=1[/TEX]
vậy N=1:cool:mình làm tắt quá mong thông cảm,nhớ thank nha!
 
H

hiendang241

bài 2 đây(thay abc=1 nha)

a/ab+a+1 +b/bc+b+1 +c/ac+c+1
=a/ab+a+abc +b/bc+b+1 +c/ac+c+abc
=1/b+1+bc +b/bc+b+1 +1/a+1+ab
=b+1/bc+1+b +abc/a+abc+ab
=b+1/bc+1+b +bc/1+bc+b
=b+1+bc/bc+1+bc
=1
 
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