Câu hỏi của Nguyễn Việt Bách

Question

Tính A = $\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{10^2}-1\right)$

Nguyễn Lê Phước Thịnh · Accepted Answer

$A=\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)\cdot...\cdot\left(\dfrac{1}{10^2}-1\right)$$=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\cdot\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)$$=\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{11}{10}$$=\dfrac{2}{10}\cdot\dfrac{11}{3}=\dfrac{22}{30}=\dfrac{11}{15}$Câu trả lời: $A=\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)\cdot...\cdot\left(\dfrac{1}{10^2}-1\right)$$=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)\cdot...\cdot\left(\dfrac{1}{10}-1\right)\cdot\left(\dfrac{1}{3}+1\right)\left(\dfrac{1}{4}+1\right)\cdot...\cdot\left(\dfrac{1}{10}+1\right)$$=\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-9}{10}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{11}{10}$$=\dfrac{2}{10}\cdot\dfrac{11}{3}=\dfrac{22}{30}=\dfrac{11}{15}$

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