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\(1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}...1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}...\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}...\frac{9.9}{8.10}\)
\(=\frac{2.3.4...9}{1.2.3...8}.\frac{2.3.4...9}{3.4.5...10}\)
\(=9.\frac{2}{10}\)
\(=9.\frac{1}{5}=\frac{9}{5}\)
1 và 1 phần 3 . 1 và 1 phần 8 . 1 và 1 phần 15 . 1 và 1 phần 24 . 1 và 1 phần 35 . 1 và 1 phần 48 . 1 và 1 phần 63 . 1 và 1 phần 80
= 4 phần 3 . 9 phần 8 . 16 phần 15 . 25 phần 24 . 36 phần 35 . 49 phần 48 . 64 phần 63 . 81 phần 80
= 3 phần 2 . 10 phần 9 . 15 phần 14 . 36 phần 35
= 5 phần 3 . 54 phần 49
= 90 phần 49
a) \(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}+\frac{1}{143}+\frac{1}{195}\)
\(=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{15}\right)\)
\(=\frac{1}{2}.\frac{14}{15}\)
\(=\frac{14}{30}=\frac{7}{15}\)
a)
\(=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}\)
\(=2\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\right)\)
\(=2\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(=2\left(1-\frac{1}{15}\right)\)
\(=2.\frac{14}{15}\)
\(=\frac{28}{15}\)
b)
\(=1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+\frac{2}{90}+\frac{2}{110}+\frac{2}{132}\)
\(=1+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+\frac{2}{9.10}+\frac{2}{10.11}+\frac{2}{11.12}\)
\(...\)
a, \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)\)
Ta có: \(\frac{1}{13}< \frac{1}{12};\frac{1}{14}< \frac{1}{12};\frac{1}{15}< \frac{1}{12}\Rightarrow\frac{1}{13}+\frac{1}{14}+\frac{1}{15}< \frac{1}{12}+\frac{1}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
\(\frac{1}{61}< \frac{1}{60};\frac{1}{62}< \frac{1}{60};\frac{1}{63}< \frac{1}{60}\Rightarrow\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{60}+\frac{1}{60}+\frac{1}{60}=\frac{3}{60}=\frac{1}{20}\)
\(\Rightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
Vậy...
b, Đặt A là tên của tổng trên
Ta có: \(A=\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{100^2}=\frac{1}{2^2}\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B là biêu thức trong ngoặc
Ta có: \(1=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow B< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow B< 1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow B< 2-\frac{1}{50}< 2\)
Thay B vào A ta được:
\(A< \frac{1}{2^2}.2=\frac{1}{2}\)
1/
a) \(C=\frac{4}{7}.\frac{3}{5}.\frac{7}{4}.\left(-20\right).\frac{5}{6}\)
\(=\left(\frac{4}{7}.\frac{7}{4}\right).\left(\frac{3}{5}.\frac{5}{6}\right).\left(-20\right)\)
\(=\frac{1}{2}.\left(-20\right)\)
\(=-10\)
2/ \(B=\frac{2^2}{3}.\frac{3^2}{8}.\frac{4^2}{15}.\frac{5^2}{24}.\frac{6^2}{35}.\frac{7^2}{48}.\frac{8^2}{63}.\frac{9^2}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}.\frac{7.7}{6.8}.\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.3.4.5.6.7.8.9}{1.2.3.4.5.6.7.8}.\frac{2.3.4.5.6.7.8.9}{3.4.5.6.7.8.9.10}\)
\(=9.\frac{2}{10}=9.\frac{1}{5}=\frac{9}{5}\)
a: \(=\left(-\dfrac{25}{140}+\dfrac{245}{140}+\dfrac{32}{140}\right)\cdot\dfrac{-69}{20}\)
\(=\dfrac{252}{140}\cdot\dfrac{-69}{20}\)
\(=\dfrac{9}{5}\cdot\dfrac{-69}{20}=\dfrac{-621}{100}\)
b: \(=\left(6-2-\dfrac{4}{5}\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}=\dfrac{18}{5}\)
c: \(=\left(\dfrac{2}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{34}{24}\cdot\dfrac{-8}{17}=\dfrac{-1}{3}\cdot2=-\dfrac{2}{3}\)
a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)
\(\Rightarrow82< x< 92\)
\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)
b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)
\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)
\(\Rightarrow-1\le x\le7\)
\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)