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=(1975/1976+2010/2011+1963/1968)x(4/12-3/12-1/12)
=(1975/1976+2010/2011+1963/1968)x0
=0
1-2+3-4+5-6+...+2011-2012
=2012-2011+...+6-5+4-3+2-1
=(2012-2001)+...+(6-5)+(4-3)+(2-1)
=1+1+1+...+1+1(có 1006 số 1)
=1x60
=60
\(\frac{1}{2.5}\)\(+\)\(\frac{1}{5.8}\)\(+\frac{1}{8.11}\)\(+...+\frac{1}{152.155}\)
=\(\frac{1}{2}\) \(-\frac{1}{5}\) \(+\frac{1}{5}\) \(-\frac{1}{8}\) \(+...+\frac{1}{152}\) \(-\frac{1}{155}\)
=\(\frac{1}{2}\)\(-\frac{1}{155}\)
=\(\frac{153}{310}\)
a) \(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)....\left(\frac{1}{1000}-1\right)=-\frac{1}{2}.\left(-\frac{2}{3}\right).\left(-\frac{3}{4}\right)...\left(-\frac{999}{1000}\right)\)
\(=-\frac{1.2.3...999}{2.3.4...1000}=-\frac{1}{1000}\)
b)\(B=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}:\frac{3}{4}=\frac{3\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{11}\right)}:\frac{3}{4}=\frac{3}{4}:\frac{3}{4}=1\)
d) \(D=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
=> \(D=2-\frac{1}{1024}=\frac{2047}{1024}\)
Bài 2:1-2+3-4+...+2011-2012
=1+2+3+4+...+2011+2012-2(2+4+6+...+2012)
=2025078-2(1012036)
=2025078-2024072
=1006
Học giỏi!
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
Ta có : 2015 + 2013 + 2011 + ... + 5 + 3 + 1
= [(2015 - 1) : 2 + 1].(2015 + 1) : 2
= 1008.2016 : 2 = 1016064
Lại có : 2015 - 2013 + 2011 - 2009 + ... + 7 - 5 + 3 - 1 (1008 số hạng
= (2015 - 2013) + (2011 - 2009) + ... + (7 - 5) + (3 - 1) (504 cặp)
= 2 + 2 + ... + 2 + 2 (504 số hạng 2)
= 2 x 504 = 1008
Khi đó A = \(\frac{1016064}{1008}=1008\)
b) tTa có : B = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{1}{1.99}+\frac{1}{3.97}+\frac{1}{5.95}+...+\frac{1}{97.3}+\frac{1}{99.1}}\)
=> \(\frac{B}{100}\) = \(\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{\frac{100}{1.99}+\frac{100}{3.97}+\frac{100}{5.95}+...+\frac{100}{97.3}+\frac{100}{99.1}}\)
\(=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{1+\frac{1}{99}+\frac{1}{3}+\frac{1}{97}+\frac{1}{5}+\frac{1}{95}+..+\frac{1}{97}+\frac{1}{3}+\frac{1}{99}+1}=\frac{1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}}{2\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{97}+\frac{1}{99}\right)}=\frac{1}{2}\)
Khi đó : B/100 = 1/2
=> B = 50
Vậy B = 50
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
\(\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).0\)
\(=0\)
\(\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).0\)
\(=0\)