Question

[superman]

Tính bằng phương pháp hợp lí

a> C = \(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{73\cdot76}\)                b> D = \(\frac{5}{10\cdot11}+\frac{5}{11\cdot12}+\frac{5}{12\cdot13}+...+\frac{5}{99\cdot100}\)

 NHANH LÊN MÌNH CẦN GẤP

Answer 1

[KAl(SO4)2·12H2O]

a) \(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)

\(C=1.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)

\(C=1.\left(\frac{1}{4}-\frac{1}{76}\right)\)

\(C=1.\frac{9}{38}\)

\(C=\frac{9}{38}\)

b) \(D=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{99.100}\)

\(D=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\right)\)

\(D=5.\left(\frac{1}{10}-\frac{1}{100}\right)\)

\(D=5.\frac{9}{100}\)

\(D=\frac{99}{20}\)