Question

[Nguyễn Trung Hiếu]

Tính : \(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-\frac{1}{5.3}-\frac{1}{3.1}=................\)  (P/S tối giản)

CTV giỏi vô đây giải bài này hộ tui cái, cả thầy và cô nữa. Ai giải được sẽ được 2 tick.

Answer 1

[Nguyễn Ngọc An]

\(\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-\frac{1}{5.3}-\frac{1}{3.1}\)

\(=\frac{1}{2}.\left(\frac{2}{99.97}-\frac{2}{97.95}-\frac{2}{95.93}-\frac{2}{5.3}-\frac{2}{3.1}\right)\)

\(=\frac{1}{2}.\left(\frac{99-97}{99.97}-\frac{97-95}{97.95}-\frac{95-93}{95.93}-\frac{5-3}{5.3}-\frac{3-1}{3.1}\right)\)

\(=\frac{1}{2}.\left[\left(\frac{99}{99.97}-\frac{97}{99.97}\right)-\left(\frac{97}{97.95}-\frac{95}{97.95}\right)-\left(\frac{95}{95.93}-\frac{93}{95.93}\right)-\left(\frac{5}{5.3}-\frac{3}{5.3}\right)-\left(\frac{3}{3.1}-\frac{1}{3.1}\right)\right]\)

\(=\frac{1}{2}.\left[\left(\frac{1}{97}-\frac{1}{99}\right)-\left(\frac{1}{95}-\frac{1}{97}\right)-\left(\frac{1}{93}-\frac{1}{95}\right)-\left(\frac{1}{3}-\frac{1}{5}\right)-\left(\frac{1}{1}-\frac{1}{3}\right)\right]\)

\(=\frac{1}{2}.\left[\frac{1}{97}-\frac{1}{99}-\frac{1}{95}+\frac{1}{97}-\frac{1}{93}+\frac{1}{95}-\frac{1}{3}+\frac{1}{5}-\frac{1}{1}+\frac{1}{3}\right]\)

\(=\frac{1}{2}.\left[-\frac{1}{99}-\frac{1}{93}+\frac{1}{5}-\frac{1}{1}\right]\)

Answer 2

[bảo anh]

A=-(1/1.3+1/3.5+1/93.95+1/95.97+1/97.99)

A=-1/2.(2/1.3+2/3.5+2/93.95+2/95.97+2/97.99)

A=-1/2.(1/1.3+1/3.5+1/93.95+1/95.97+1/97.99)

A=-1/2(1-1/93-1/99)

A=-3005/6138

mik ko bit co dung ko nua