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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
Sửa đề:
\(\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(=\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\dfrac{4-3-1}{12}\)
\(=\left(\dfrac{1975}{1976}+\dfrac{2010}{2011}+\dfrac{1963}{1968}\right)\times\dfrac{0}{12}\)
\(=0\)
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
\(A=\frac{2015+2013+2011+...+5+3+1}{2015-2013+2011-2009+...+7-5+3-1}\)
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
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Ta có: \(C=\frac{\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}}\)
Đặt \(A=\frac{2010}{1}+\frac{2009}{2}+\frac{2008}{3}+...+\frac{1}{2010}\)
\(A=\frac{2010}{1}+1+\frac{2009}{1}+1+\frac{2008}{1}+1+...+\frac{1}{2010}+1-2010\)
\(=\frac{2011}{1}+\frac{2011}{2}+\frac{2011}{3}+...+\frac{2011}{2010}-\frac{2011.2010}{2011}\)
\(=2011\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\right)\)
Đặt \(B=\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}\)
\(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}-1\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}-\frac{2010}{2011}\)
Ta có: \(C=\frac{A}{B}=2011\)(lấy A-B)
Ta có :
\(2010A=\dfrac{2010^{2012}+2010}{2010^{2012}+1}=\dfrac{2010^{2012}+1+2009}{2010^{2012}+1}=1+\dfrac{2009}{2010^{2012}+1}\)
\(2010B=\dfrac{2010^{2011}+2010}{2010^{2011}+1}=\dfrac{2010^{2011}+1+2009}{2010^{2011}+1}=1+\dfrac{2009}{2010^{2011}+1}\)
Vì \(1+\dfrac{2009}{2010^{2012}+1}< 1+\dfrac{2009}{2010^{2011}+1}\Rightarrow A< B\)
~ Học tốt ~
\(\frac{1}{2.5}\)\(+\)\(\frac{1}{5.8}\)\(+\frac{1}{8.11}\)\(+...+\frac{1}{152.155}\)
=\(\frac{1}{2}\) \(-\frac{1}{5}\) \(+\frac{1}{5}\) \(-\frac{1}{8}\) \(+...+\frac{1}{152}\) \(-\frac{1}{155}\)
=\(\frac{1}{2}\)\(-\frac{1}{155}\)
=\(\frac{153}{310}\)
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) \(D=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{512}+\frac{1}{1024}\)
=> \(2D=2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\)
=> \(2D-D=\left(2+1+\frac{1}{2}+...+\frac{1}{256}+\frac{1}{512}\right)-\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}+\frac{1}{1024}\right)\)
=> \(D=2-\frac{1}{1024}=\frac{2047}{1024}\)
\(\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).0\)
\(=0\)
\(\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{1}{3}-\frac{1}{4}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).\left(\frac{4}{12}-\frac{3}{12}-\frac{1}{12}\right)\)
\(=\left(\frac{1975}{1976}+\frac{2010}{2011}+\frac{1963}{1968}\right).0\)
\(=0\)