Question

[Hii Văn]

Bài: Tính:

P=\(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{2.1}\).

M=\(\frac{3}{2}+\frac{5}{4}+\frac{9}{8}+\frac{17}{16}+\frac{33}{32}+\frac{65}{64}-7\)

Q=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{64}\)

E=\(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\frac{1}{12}\)

F=\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{62.65}\)

Answer 1

[★๖ۣۜßảo๖ۣۜPɦα♏๖ۣۜ[EηgĻïšħ☯€lub]★]

\(P=...\)

\(=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+\frac{1}{97}-...-\frac{1}{2}+1\)

\(=\frac{1}{99}-1=\frac{-98}{99}\)

Answer 2

[★๖ۣۜßảo๖ۣۜPɦα♏๖ۣۜ[EηgĻïšħ☯€lub]★]

\(M=...\)

\(=\frac{2}{2}+\frac{1}{2}+\frac{4}{4}+\frac{1}{4}+...+\frac{64}{64}+\frac{1}{64}-7\)

\(=1+1+1+1+1+1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+\frac{1}{2^6}-7\)

\(=\frac{1+2+2^2+2^3+2^4+2^5}{2^6}-1\)

\(=\frac{2^6-1}{2^6}-1=1-\frac{1}{2^6}-1=-\frac{1}{2^6}\)