Question

[vampire ntn]

tính rồi so sánh m và n biết 

m = \(\frac{3}{1\cdot3}\)+ \(\frac{3}{3\cdot5}\)+\(\frac{3}{5\cdot7}\)+...+\(\frac{2}{99\cdot100}\)

n = \(\frac{3}{1\cdot3}\)+ \(\frac{3}{3\cdot5}\)+ \(\frac{3}{5\cdot7}\)+ ...+ \(\frac{3}{97\cdot99}\)

rồi, mấy bạn giải ra giùm mk nhé. 4h mk di học rồi cho nên trước 4 giờ các bạn giải cho mk nhé.

Answer 1

[Sáng Nguyễn]

n=\(\frac{2}{3}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)

n=\(\frac{2}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

n=\(\frac{2}{3}\left(1-\frac{1}{99}\right)\)

n=\(\frac{2}{3}\times\frac{98}{99}\)

n=\(\frac{196}{297}\)

Answer 2

[Đức Phạm]

Câu \(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{2}{99.100}\)Bạn viết \(\frac{3}{99.100}=\frac{2}{99.100}\)mik sửa lại nhé. 

\(M=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{99.100}\)

\(M=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+...+\frac{100-99}{99.100}\)

\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)

\(M=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{100}\right)\)

\(M=\frac{3}{2}.\frac{99}{100}=\frac{297}{200}\)

\(N=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+....+\frac{3}{97.99}\)

\(N=\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+....+\frac{99-97}{97.99}\)

\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{97}-\frac{1}{99}\right)\)

\(N=\frac{3}{2}.\left(\frac{1}{1}-\frac{1}{99}\right)\)

\(\Rightarrow N=\frac{3}{2}.\frac{98}{99}=\frac{49}{33}\)

Ta thấy : \(\frac{297}{200}>\frac{49}{33}\Rightarrow M>N\)