cho\(A=\frac{1}{1.2}+\frac{1}{3.4}+.........+\frac{1}{99.100}CMR\frac{7}{12}< A< \frac{5}{6}\)
Trước hết ta biến đổi A thành \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Do đó : \(A=\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right]+\left[\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right]\)
Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75},\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên
\(A>\frac{1}{75}\cdot25+\frac{1}{100}\cdot25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(A< \frac{1}{51}\cdot25+\frac{1}{76}\cdot25< \frac{1}{50}\cdot25+\frac{1}{75}\cdot25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy \(\frac{7}{12}< A< \frac{5}{6}\)
Cách biến đổi :
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
Dễ thấy \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right]\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}-\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right]\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
CMR: \(\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{80}>\frac{ }{ }\)7/12
B=\(\frac{1}{101}+\frac{1}{102}+......+\frac{1}{200}\)CMR : B >\(\frac{7}{12}\)
\(B=\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{150}\right)+\left(\frac{1}{151}+\frac{1}{152}+...+\frac{1}{200}\right)>50.\frac{1}{150}+50.\frac{1}{200}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(A=\frac{1}{1\times2}+\frac{1}{3\times4}+\frac{1}{5\times6}+...+\frac{1}{99\times100}.CMR:\frac{7}{12}< A< \frac{5}{6}\)
CMR:
\(\frac{7}{12}< \frac{1}{41}+\frac{1}{42}+\frac{1}{43}+.........+\frac{1}{79}+\frac{1}{80}< \frac{5}{6}\)
Đặt \(A=\frac{1}{41}+\frac{1}{42}+\frac{1}{43}+...+\frac{1}{79}+\frac{1}{80}\)
\(A>\frac{1}{80}+\frac{1}{80}+....+\frac{1}{80}\)
\(\frac{1}{80}+\frac{1}{80}+.....+\frac{1}{80}\\ =\frac{40}{80}=\frac{1}{2}\)
Vì \(\frac{1}{2}< \frac{5}{6}\\ =>A< \frac{5}{6}\)
\(A< \frac{1}{40}+\frac{1}{40}+.....+\frac{1}{40}\)
\(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\\ =\frac{40}{40}=1\)
Vì \(1>\frac{7}{12}\\ =>A>\frac{7}{12}\)
bài này đề có vấn để
\(ChoA=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}CMR\frac{7}{12}
cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
CMR \(\frac{7}{12}< A< \frac{5}{6}\)
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mik ghi từ trong sách mà
Cho \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{99.100}\). CMR: \(\frac{7}{12}< A< \frac{5}{6}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+............+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A=1-\frac{1}{100}\)
\(\Rightarrow A=\frac{99}{100}\)
Vì \(\frac{7}{12}< \frac{99}{100}< \frac{5}{6}\Rightarrow\frac{7}{12}< A< \frac{5}{6}\) ĐPCM
( Bài này ko ai lm thì t lm cho 
)
Cho A = \(\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\). CMR: \(\frac{7}{12}< A< \frac{5}{6}\)
Câu hỏi của Vũ Thị Kim Oanh - Toán lớp 7 - Học toán với OnlineMath
\(choA=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(CMR:\frac{7}{12}< A< \frac{5}{6}\)
