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tungnguyen14711
2 năm trước
Tính B=$\frac{2.1+1}{\left[1\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2\left(2+1\right)\right]^2}+\frac{2.3+1}{\left[3\left(3+1\right)\right]^2}+.. $\frac{2.99+1}{[99(99+1)]^{2}}$
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