help me

N

nguyenbahiep1

câu b)

[TEX]\frac{1}{5.8} + \frac{1}{8.11}+.....+ \frac{1}{x(x+3)} = \frac{101}{1540} \\ \frac{1}{3}(\frac{1}{5} - \frac{1}{8} + \frac{1}{8}-\frac{1}{11}+....+\frac{1}{x}-\frac{1}{x+3}) = \frac{101}{1540} \\ \frac{1}{3}(\frac{1}{5}-\frac{1}{x+3}) = \frac{101}{1540} \\ \frac{1}{5}-\frac{1}{x+3} = \frac{303}{1540} \\ \frac{1}{x+3} = \frac{1}{308} \\ \Rightarrow x = 305[/TEX]
 
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