toán 8 nâng cao đây

K

khanhtoan_qb

Cho abc=1 tính
[TEX]A =\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+ 1}[/TEX]
Ta có :
ab + a + 1 = ab + a + abc = a(bc + b + 1) = a(bc + b + abc) = ab(ac + c + 1)
\Rightarrow [TEX]bc + b + 1 = \frac{ab + a + 1}{a}[/TEX]
và [TEX]ac + c + 1 = \frac{ab + a + 1}{ab}[/TEX]
\Rightarrow[TEX]A = \frac{a}{ab + b + 1} + \frac{ab}{ab + b + 1} + \frac{abc}{ab + b + 1}[/TEX]
\Rightarrow[TEX]A = \fra{a + ab + abc}{ab + a + 1} [/TEX]
\Rightarrow[TEX]A = \frac{ab + a + 1}{ab + a + 1} = 1[/TEX](*)(*)(*)
 
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