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\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
a/ (-3,2).\(\frac{-15}{64}\)+(0,8-2\(\frac{4}{5}\)):1\(\frac{23}{24}\)
=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(\(\frac{4}{5}\)-\(\frac{14}{5}\)):\(\frac{47}{24}\)
=(\(\frac{-16}{5}\)).\(\frac{-15}{64}\)+(-2):\(\frac{47}{24}\)
= \(\frac{3}{4}\)+\(\frac{-48}{47}\)
=\(\frac{-51}{188}\)
b/ 1\(\frac{13}{15}\).3.(0,5)\(^2\).3+(\(\frac{8}{15}\)-1\(\frac{19}{60}\)):1\(\frac{23}{24}\)
= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{8}{15}\)-\(\frac{79}{60}\)):\(\frac{47}{24}\)
= \(\frac{28}{15}\).3.\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{28}{5}\).\(\frac{1}{4}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{7}{5}\).3+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{21}{5}\)+(\(\frac{-47}{60}\)):\(\frac{47}{24}\)
= \(\frac{21}{5}\)+(\(\frac{-2}{5}\))
= \(\frac{19}{5}\)
mk làm hơi dài dòng chút
CHÚC BẠN HỌC TỐT
\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)
=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)
=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)
Đáp số: C=1
a) C=\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{999}{1000}\)\(=\frac{1.2.3...999}{2.3.4...1000}=\frac{1.\left(2.3.4....999\right)}{\left(2.3.4....999\right).1000}\)\(=\frac{1}{1000}\)
b) Đặt: A=\(1+2+2^2+2^3+...+2^{2008}\)
\(\Leftrightarrow2A=2+2^2+2^3+....+2^{2008}+2^{2009}\)
\(\Leftrightarrow2A-A=2^{2009}-1\)
\(\Leftrightarrow A=2^{2009}-1\)
\(\Rightarrow S=\frac{2^{2009}-1}{1-2^{2009}}\)\(=\frac{2^{2009}-1}{-\left(2^{2009}-1\right)}=\frac{1}{-1}=-1\)
vậy: S=(-1)