A=\(\frac{1}{15}+\frac{1}{16}+...+\frac{1}{44}\)
CMR: A >\(\frac{5}{6}\)
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So sanh A va B:
\(A=\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}\)
\(B=\frac{5}{6}\)
Chứng minh rằng
\(\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{43}+\frac{1}{44}>\frac{5}{6}\)
Bài 1:CMR:\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{15}<2\)
Bài 2: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right)}=\frac{99}{101}\)
Bài 3:\(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2449}{2500}\)
Bài 4:CMR:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{49}-\frac{1}{50}=\frac{1}{26}+\frac{1}{27}+\frac{1}{28}+...+\frac{1}{50}\)
1/ CMR: -a + 3 và 3 - a là 2 số đối nhau
2/ Cho C =\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}\). Chứng tỏ rằng C<2
3/ CMR:
a) \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}<1\)
b) \(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}<\frac{1}{2}\)
1. tìm tích của A= \(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times..\times\frac{899}{900}\)
2. CMR \(\frac{1}{5}+\frac{1}{6}+..+\frac{1}{17}< 2\)
3. tính \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+..+\frac{1}{10.11.12}\)
3. \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\)
\(\Leftrightarrow2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{10.11.12}\)
\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{11.12}\)
\(\Leftrightarrow2M=\frac{1}{2}-\frac{1}{132}\)
\(\Leftrightarrow2M=\frac{65}{132}\)
\(\Leftrightarrow M=\frac{65}{132}\div2\)
\(\Leftrightarrow M=\frac{65}{264}\)
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1\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)
\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)
\(\Leftrightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)
\(\Leftrightarrow A=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4...30\right)\left(2.3.4...30\right)}\)
\(\Leftrightarrow A=\frac{1.31}{30.2}\)
\(\Leftrightarrow A=\frac{31}{60}\)
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2. Đặt \(A=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)
\(\Rightarrow A< \frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\)
\(\Rightarrow A< 1+1=2\)
Vậy a < 2 (đpcm)
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Chứng minh rằng
\(\frac{1}{5}+\frac{1}{16}+\frac{1}{17}+...+\frac{1}{44}+\frac{1}{45}>\frac{5}{6}\)
CMR :a) \(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+.....+\frac{1}{100^2}< \frac{1}{2}\)
b) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+....+\frac{2499}{2500}>48\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)\)
TA có :\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}=1-\frac{1}{50}\)
=>\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1\Rightarrow1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+1=2\)
\(A=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}\right)< \frac{1}{2^2}.2=\frac{1}{2}\left(đpcm\right)\)
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\(ChoA=\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}CMR\frac{7}{12}
Tính:
\(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+.....+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2017}.\)
\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}.\)