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\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(A=\frac{2}{5.7}+\frac{4}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+\frac{1}{17}-\frac{1}{19}+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\)
\(A=\frac{1}{5}-\frac{1}{25}=\frac{4}{25}\)
\(A=\frac{2}{35}+\frac{4}{77}+\frac{2}{143}+\frac{4}{221}+\frac{2}{323}+\frac{4}{437}+\frac{2}{575}\)
\(A=\frac{2}{5.7}+\frac{4}{7.11}+\frac{2}{11.13}+\frac{4}{13.17}+\frac{2}{17.19}+\frac{4}{19.23}+\frac{2}{23.25}\)
\(A=1.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+...+\frac{1}{19}-\frac{1}{23}+\frac{1}{23}-\frac{1}{25}\right)\)
A = 1/5 - 1/25
A = 4/25
13)\(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}\)
\(=1-\frac{1}{9}\)
\(=\frac{8}{9}\)
Các bạn , mk xin lỗi nha, cái câu 16 ấy đáng lẽ phải như vầy:
a) A= 1/2010+1+2/2009+1+3/2008+1+...+2009/2+1+1
= 2011/2010+20011/2009+2011/2008+...+2011/2+2011/2011
= 2011(1/2+1/3+1/4+...+1/2011)
Ta có: B= 1/2+1/3+1/4+...+1/2011
suy ra A/B= 2011
\(B=\frac{4}{35}+\frac{4}{63}+\frac{4}{99}+\frac{4}{143}+\frac{4}{195}\)
\(B=2\cdot\left(\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}+\frac{2}{13\cdot15}\right)\)
\(B=2\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{15}\right)\)
\(B=2\cdot\left(\frac{1}{5}-\frac{1}{15}\right)\)
\(B=2\cdot\frac{2}{15}\)
\(B=\frac{4}{15}\)
B = 4/35 + 4/63 + 4/99 + 4/143 + 4/195
B = 4/5.7 + 4/7.9 + 4/9.11 + 4/11.13 + 4/13.15
B = 4/2 . ( 2/5.7 + 2/7.9 + 2/9.11 + 2/11.13 + 2/13.15 )
B = 4/2 . ( 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11 + 1/11 - 1/13 + 1/13 - 1/15 )
B = 4/2 . ( 1/5 - 1/15 )
B = 4/2 . 2/15
B = 4/15
Bạn giải cũng được đấy alibaba nguyễn, nhưng theo mình thì làm cách này dễ hiểu hơn!
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 \(1+\dfrac{2009}{2010^{2012}+1}< 1+\dfrac{2009}{2010^{2011}+1}\Rightarrow A< B\)
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