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Question

$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}$

OH-YEAH^^ · Accepted Answer

Đặt A=$\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}$$3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}$$3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}$$3A=\dfrac{1}{2}-\dfrac{1}{98}$$3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}$Câu trả lời: Đặt A=$\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}$$3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}$$3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}$$3A=\dfrac{1}{2}-\dfrac{1}{98}$$3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}$

Phía sau một cô gái · Answer

$\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}$$=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}$$=\dfrac{1}{2}-\dfrac{1}{98}$$=\dfrac{24}{49}$Câu trả lời:     $\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}$$=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}$$=\dfrac{1}{2}-\dfrac{1}{98}$$=\dfrac{24}{49}$

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