Help: Cho A=\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}\).Hãy so sánh A với \(\frac{1}{2}\)
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Hãy giúp tôi giải phép tính này:
So sánh:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2}+\frac{1}{11^2}\)với\(\frac{5}{12}\)
Ta có
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{11^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}\)
Mà
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{11.12}=\frac{3-2}{2.3}+\frac{4-3}{3.4}\frac{5-4}{4.5}+...+\frac{12-11}{11.12}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{11}-\frac{1}{12}=\frac{1}{2}-\frac{1}{12}=\frac{5}{12}\)
Nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}>\frac{5}{12}\)
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1/2^2+1/3^2+1/4^2+....+1/11^2<1/(2.3)+1/(3.4)+1/(4.5)+.....+1/(11.12)
=1/2-13+1/3-1/4+1/5+.....+1/11-1/12
=1/2-1/12=5/12
VẬY A<5/12
ks cho mình nhé
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Cho A =\(\frac{1}{10}\)+ \(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{100}\)Hãy so sánh A với \(\frac{1}{2}\)
\(A=\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+...+\frac{1}{100}\)
\(A< \frac{1}{10.11}+\frac{1}{11.12}+...+\frac{1}{100.101}\)
\(A< \frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+...+\frac{1}{100}-\frac{1}{101}\)
\(A< \frac{1}{10}-\frac{1}{101}=\frac{101}{1010}-\frac{10}{1010}=\frac{91}{1010}< \frac{505}{1010}\)
\(A< \frac{1}{2}\)
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Cho A = \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
B = \(\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{10}\right)\)
a) So sánh A và B
b) Chứng minh A = \(\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\)
bài 1: So sánh
a,\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2} và 1 \)
b,\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2} và 1\)
ai giải hộ mik vs:CMR:\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2}+\frac{1}{14^2}< \frac{1}{2}\)
26 tháng 9 2019 lúc 21:13
So sánh: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\) với 1
- Plz help me :(
Ta có : \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}\)
\(=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
\(< \frac{1}{4}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{\left(n-1\right)n}\right)\)
\(=\frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(=\frac{1}{4}.\left(2-\frac{1}{n}\right)\)
\(=\frac{1}{2}-\frac{1}{4n}< 1\)
Vậy A < 1
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\(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{\left(2n\right)^2}.\)
\(A=\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+...+\frac{1}{4n^2}.\)
\(A=\frac{1}{4}\left(1+\frac{1}{4}+\frac{1}{9}+...+\frac{1}{n^2}\right)\)
\(A=\frac{1}{4}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2}\right)\)
So sánh \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};....\)
\(\Rightarrow A< \frac{1}{4}\left(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{n\left(n-1\right)}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{n-1}+\frac{1}{n-1}-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(1+1-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{4}\left(2-\frac{1}{n}\right)\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}\)
có \(\frac{1}{2}>\frac{1}{2}-\frac{1}{4n}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{4n}< \frac{1}{2}\) mà \(\frac{1}{2}< 1\)
\(\Rightarrow A< 1\)
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So sánh A ;B : \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2};B=\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
Ta có:
\(2^2<4^2\Rightarrow\frac{1}{2^2}>\frac{1}{4^2}\)
\(3^2<6^2\Rightarrow\frac{1}{3^2}>\frac{1}{6^2}\)
\(4^2<8^2\Rightarrow\frac{1}{4^2}<\frac{1}{8^2}\)
\(...\)
\(100^2<200^2\Rightarrow\frac{1}{100^2}>\frac{1}{200^2}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}>\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+...+\frac{1}{200^2}\)
\(\Rightarrow A>B\)
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A>B vì mẫu số càng nhỏ thì phân số càng lớn
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Xem thêm câu trả lời
1. tính A frac{3}{2^2}.frac{8}{3^2}.frac{15}{4^2}...frac{899}{30^2}2. tính B frac{1}{4}.frac{2}{6}.frac{3}{8}.frac{4}{10}...frac{30}{62}.frac{31}{64}3. So sánh C left(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right)...left(1-frac{1}{20}right)với frac{1}{21}4. So sánh D left(1-frac{1}{4}right).left(1-frac{1}{9}right).left(1-frac{1}{16}right)...left(1-frac{1}{100}right)với frac{11}{19}
Đọc tiếp
1. tính A= \(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{899}{30^2}\)
2. tính B= \(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}...\frac{30}{62}.\frac{31}{64}\)
3. So sánh C= \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{20}\right)\)với \(\frac{1}{21}\)
4. So sánh D= \(\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{100}\right)\)với \(\frac{11}{19}\)
\(\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}.....\frac{899}{30^2}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4.....30}.\frac{3.4.5.....31}{2.3.4.....30}\)
\(=\frac{1}{2}.\frac{31}{30}=\frac{31}{60}\)
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
Đọc tiếp
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
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