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A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.........\frac{899}{900}\)
A=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}..........\frac{29.31}{30.30}\)
A=\(\frac{1.2.3.......29}{2.3.4.......30}.\frac{3.4.5........31}{2.3.4.......30}\)
A=\(\frac{1}{30}.\frac{2}{31}=\frac{1}{465}\)
a) \(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{302\cdot305}\)
\(B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{302\cdot305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{302}-\frac{1}{305}\right)\)
\(B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{305}\right)=\frac{1}{3}\cdot\frac{303}{610}=\frac{101}{610}\)
b) \(C=\frac{6}{1\cdot4}+\frac{6}{4\cdot7}+....+\frac{6}{202\cdot205}\)
\(C=2\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+...+\frac{3}{202\cdot205}\right)=2\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\right)\)
\(=2\left(1-\frac{1}{205}\right)=2\cdot\frac{204}{205}=\frac{408}{205}\)
c) \(D=\frac{5^2}{1\cdot6}+\frac{5^2}{6\cdot11}+...+\frac{5^2}{266\cdot271}\)
\(D=5\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+...+\frac{5}{266\cdot271}\right)\)
\(D=5\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\right)=5\left(1-\frac{1}{271}\right)=5\cdot\frac{270}{271}=\frac{1350}{271}\)
d) \(E=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{5}{16}\cdot...\cdot\frac{9999}{10000}=\frac{3\cdot8\cdot15\cdot...\cdot9999}{4\cdot9\cdot16\cdot...\cdot10000}=\frac{3}{10000}\)
e) \(F=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(F=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2500}\right)\)
\(F=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}=\frac{3\cdot8\cdot15\cdot...\cdot2499}{4\cdot9\cdot16\cdot...\cdot2500}=\frac{3}{2500}\)
a. \(B=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{302.305}\)
\(\Rightarrow3B=\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{302.305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{302}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{1}{2}-\frac{1}{305}\)
\(\Rightarrow3B=\frac{303}{610}\)
\(\Rightarrow B=\frac{101}{610}\)
b. \(C=\frac{6}{1.4}+\frac{6}{4.7}+...+\frac{6}{202.205}\)
\(\Rightarrow\frac{1}{2}C=\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{202.205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{202}-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=1-\frac{1}{205}\)
\(\Rightarrow\frac{1}{2}C=\frac{204}{205}\)
\(\Rightarrow C=\frac{408}{205}\)
c. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{889}{900}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29\cdot31}{30.30}\)
\(=\frac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4....30.30}\)
\(=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4....30\right)\left(2.3.4.....30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
có \(\frac{1}{2\cdot3}< \frac{1}{2^2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3\cdot4}< \frac{1}{3^2}< \frac{1}{2\cdot3}\)
\(\frac{1}{4\cdot5}< \frac{1}{4^2}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{9\cdot10}< \frac{1}{9^2}< \frac{1}{8\cdot9}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{8\cdot9}>A>\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(\Rightarrow1-\frac{1}{9}>A>\frac{1}{2}-\frac{1}{10}\)
\(\Rightarrow\frac{8}{9}>A>\frac{2}{5}\)
Bạn ơi, sai rồi, mình k nhầm
làm sao mà \(\frac{1}{2^2}< \frac{1}{1.2}\)được
a) \(\frac{1}{3}-\frac{-1}{6}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
b) \(2\frac{1}{3}+4\frac{1}{5}=\frac{7}{3}+\frac{21}{5}=\frac{98}{15}\)
c) \(\frac{4}{9}-\frac{13}{3}-\frac{4}{9}-\frac{10}{3}=\left(\frac{4}{9}-\frac{4}{9}\right)-\left(\frac{13}{3}+\frac{10}{3}\right)\)
\(=0-\frac{23}{3}=\frac{-23}{3}\)
d) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
A= \(\frac{1^2}{1.2}\). \(\frac{2^2}{2.3}\). \(\frac{3^2}{3.4}\). \(\frac{4^2}{5.6}\).
A= \(\frac{1.1}{1.2}\). \(\frac{2.2}{2.3}\). \(\frac{3.3}{3.4}\). \(\frac{4.4}{4.5}\).
A= \(\frac{1.2.3.4}{1.2.3.4}\). \(\frac{1.2.3.4}{2.3.4.5}\).
A= 1. \(\frac{1}{5}\).
A= \(\frac{1}{5}\).
Vậy A= \(\frac{1}{5}\).
B= \(\frac{3}{4}\). \(\frac{8}{9}\). \(\frac{15}{16}\)..... \(\frac{899}{900}\).
B= \(\frac{1.3}{2.2}\). \(\frac{2.4}{3.3}\). \(\frac{3.5}{4.4}\)..... \(\frac{29.31}{30.30}\).
B= \(\frac{1.2.3.....29}{2.3.4.....30}\). \(\frac{3.4.5.....31}{2.3.4.....30}\).
B= \(\frac{1}{30}\). \(\frac{31}{2}\).
B= \(\frac{31}{60}\).
Vậy B= \(\frac{31}{60}\).
\(A=\frac{1.3}{2.2}.\frac{2.4}{3.3}...\frac{29.31}{30.30}\)
\(=\frac{\left(1.2....29\right).\left(3.4....31\right)}{\left(2.3....30\right).\left(2.3....30\right)}\)
\(=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
A=\(\frac{1.3}{2^2}.\frac{2.4}{3^2}.......\frac{29.31}{30^2}\)
A=\(\frac{1.3.2.4.....29.31}{2.2.3.3.4.4.....30.30}\)
A=\(\frac{1.2.3....29}{2.3.4....30}\)x \(\frac{3.4....31}{2.3.4.....30}\)
A=\(\frac{1}{30}\times\frac{31}{2}\)=\(\frac{31}{60}\)
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