cho A=(1/2^2-1) (1/3^2-1) (1/4^2-1) ... (1/2013^2-1) (1/2014^2-1) và B=-1/2 .
so sanh A va B
NT
Những câu hỏi liên quan
cho A=1+1/2+1/3+1/4+.......+1/4026 va B=1+1/3+1/5+........+1/4025
so sanh A/B va 1/2013/2014
A=(1/22-1)(1/32-1)(1/42-1)...(1/20132-1)(1/20142) ; B=-1/2 . So sanh A va B
Cho \(A=\left(\frac{1}{^{2^2}}-1\right).\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right)......\left(\frac{1}{2013^2}-1\right).\left(\frac{1}{2014^2}-1\right)va\)\(B=-\frac{1}{2}.\)Hay so sanh A va B
P=1/1^2+1/2^2+1/3^2+1/4^2+.......+1/2013^2+1/2014^2
Q=1+3/4
So sanh P va Q
cho A=1+1/2+1/3+1/4+......+1/4026, B=1+1/3+1/5+1/7+...+1/4025 so sanh a/b voi1+2013/2014
Cho A = \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{2013^2}-1\right)\left(\frac{1}{2014^2}-1\right)\) va B = \(\frac{-1}{2}\), So sanh A va B
A = \(-\frac{1.3}{2.2}.-\frac{2.4}{3.3}.\cdot\cdot\cdot-\frac{2013.2015}{2014.2014}=-\frac{\left(1.2.3...2013\right).\left(3.4.5....2015\right)}{\left(2.3....2014\right).\left(2.3....2014\right)}=-\frac{2.2015}{2014}=-\frac{4030}{2014}
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1. So sanh:
2014×2015-2/2013+2013×2014 voi 2014×2015-1/2014×2015
2. Cho a, b, c thuoc N* va a nho hon b.
Hay chung to: a/b nho hon a+c/b+c va 1 nho hon a/a+b +b/b+c+c/a+c
Cho A bang 1+1/2+1/2^2+1/2^3+1/2^4+......+1/2^2013.Hay so sanh A va 2
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\)
=>\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
=>\(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2013}}\right)\)
=>\(A=2-\frac{1}{2^{2013}}< 2\)
Vậy A<2
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Cho A=(1/2^2-1)x(1/3^2-1)x(1/4^2-1)...(1/2013^2-1)x(1/2014^2-1) và B=-1/2.so sánh A và B
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)\left(\frac{1}{4^2}-1\right)...\left(\frac{1}{2014^2}-1\right)\)
\(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2014^2}\right)\)
\(-A=\frac{3}{2\cdot2}\cdot\frac{8}{3\cdot3}\cdot\frac{15}{4\cdot4}\cdot...\cdot\frac{4056195}{2014\cdot2014}\)
\(-A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2014\cdot2014\right)}\)
\(-A=\frac{\left(1\cdot2\cdot3\cdot...\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2014\right)}\)
\(-A=\frac{1\cdot2015}{2014\cdot2}=\frac{2015}{4028}\)
\(A=\frac{-2015}{4028}\)
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