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k chép đề
3/2.A=\(\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^5+...+\left(\frac{3}{2}\right)^{2013}\)
3/2A-A=(\(\frac{3}{4}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^5+...+\left(\frac{3}{2}\right)^{2013}\)) - (\(\frac{1}{2}+\frac{3}{2}+\left(\frac{3}{2}\right)^2+\left(\frac{3}{2}\right)^3+\left(\frac{3}{2}\right)^4+...+\left(\frac{3}{2}\right)^{2012}\))
1/2 . A =\(\frac{1}{2}+\left(\frac{3}{2}\right)^{2013}\)
A=\(\frac{\frac{1}{2}+\left(\frac{3}{2}\right)^{2013}}{2}\)
B-A=\(\frac{\left(\frac{3}{2}\right)^{2018}}{2}-\)\(\frac{\frac{1}{2}+\left(\frac{3}{2}\right)^{2013}}{2}\)
\(B-A=\frac{\frac{1}{2}}{2}=\frac{1}{2}:2=\frac{1}{4}\)
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. \(D=\frac{5^2}{1.6}+\frac{5^2}{6.11}+...+\frac{5^2}{266.271}\)
\(\Rightarrow\frac{1}{5}D=\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{266.271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{266}-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=1-\frac{1}{271}\)
\(\Rightarrow\frac{1}{5}D=\frac{270}{271}\)
\(\Rightarrow D=\frac{1350}{271}\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{9^2}\)
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
=\(\left(4-2+3\right)\cdot\frac{-1}{2}\)
=\(5\cdot\left(\frac{-1}{2}\right)\)
=\(\frac{5\cdot\left(-1\right)}{2}\)
=\(\frac{-6}{2}\)
\(=\left(-3\right)\)
Có 2 trg hợp nhé: Nếu x là dấu nhân thì thực hiện theo phép nhân
Nếu x là ẩn số thì ko làm đc nhé vì ko có kết quả
Nên làm theo trường hợp 1
\(4.\frac{-1}{2}-2.\frac{-1}{2}+3.\frac{-1}{2}\)\(=\)\(\left(\frac{-1}{2}\right).\left(4-2+3\right)=\left(\frac{-1}{2}\right).5=\frac{-1.5}{2}=\frac{-5}{2}\)
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) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(< =>2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(< =>2A-A=1-\frac{1}{2^{99}}< =>A=1-\frac{1}{2^{99}}\)
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{99}}\)
\(\Rightarrow A=1-\frac{1}{2^{99}}\)