toán 6 tìm x siêu cấp

M

minhnlyb2001

Tìm x
($\frac{11}{12}+\frac{11}{12.23}+\frac{11}{23.24}+...+\frac{11}{99.100}$)
Đọc kĩ đi anh bạn:
=(11/12+11/12.23)+(11.(1/23.24+1/24.25+...+1/99.100)
=2/23+11.(1/23-1/24+1/24-1/25+...+1/99-1/100)
=2/23+11.(1/23-1/100)
=2/23+11.77/2300
=2/23+847/2300
=1047/2300
 
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