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\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
\(D=\dfrac{\left(2!\right)^2}{1^2}+\dfrac{\left(2!\right)^2}{3^2}+\dfrac{\left(2!\right)^2}{5^2}+...+\dfrac{\left(2!\right)^2}{2015^2}\)
\(D=\left(2!\right)^2\left(\dfrac{1}{3^2}+\dfrac{1}{5^2}+...+\dfrac{1}{2015^2}\right)\)
Xét số hạng tổng quát dạng: \(\dfrac{1}{\left(2n+1\right)^2}\) với \(n\in N\ge1\)
Ta có: \(\left(2n+1\right)^2-2n\left(2n+1\right)=1>0\)
\(\Rightarrow\left(2n+1\right)^2>2n\left(2n+1\right)\Rightarrow\dfrac{1}{\left(2n+1\right)^2}< \dfrac{1}{2n\left(2n+1\right)}\)
Do đó: \(\left\{{}\begin{matrix}\dfrac{1}{3^2}< \dfrac{1}{2.4}\\\dfrac{1}{5^2}< \dfrac{1}{4.6}\\....\\\dfrac{1}{2015^2}< \dfrac{1}{2014.2016}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{1^2}+\dfrac{1}{3^2}+\dfrac{1}{5^2}...+\dfrac{1}{2015^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow\dfrac{D}{\left(2!\right)^2}< 1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+..+\dfrac{1}{2014.2016}\)
\(\Leftrightarrow D< 4\left(1+\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{2014.2016}\right)\)
\(\Leftrightarrow D< 4+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{1007.1008}\)
\(\Leftrightarrow D< 4+\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{1008-1007}{1007.1008}\)
\(\Leftrightarrow D< 4+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{...1}{1007}-\dfrac{1}{1008}\)
\(\Leftrightarrow D< 5-\dfrac{1}{1008}< 5< 6\)
Bài 2:
Ta có: \(11^{1979}< 11^{1980}=1331^{660}\)
\(37^{1320}=37^{2\cdot660}=1369^{660}\)
mà \(1331^{660}< 1369^{660}\)
nên \(11^{1979}< 37^{1320}\)
a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3
=>x=-1/3+3/4=-4/12+9/12=5/12
b: =>x(1/2-5/6)=7/2
=>-1/3x=7/2
hay x=-21/2
c: (4-x)(3x+5)=0
=>4-x=0 hoặc 3x+5=0
=>x=4 hoặc x=-5/3
d: x/16=50/32
=>x/16=25/16
hay x=25
e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4
=>2x=-7/4+3=5/4
hay x=5/8
`A = 3/4 xx 8/9 xx ... xx 99/100`
`= (1xx3)/(2xx2) xx (2xx4)/(3xx3) xx ... xx (9xx11)/(10xx10)`
`= (1xx2xx3xx ... xx 9)/(2xx3xx...xx10) xx (3xx4xx5xx...xx 11)/(2xx3xx4xx...xx 10)`
`= 1/10 xx 11`
`= 11/10`.
Ta có: `11/10 > 1`
`11/19 < 1`.
`=> A > 11/19`.
Ta có: \(\left(-\dfrac{1}{5}\right)^{30}=\left[\left(-\dfrac{1}{5}\right)^3\right]^{10}=\left(-\dfrac{1}{125}\right)^{10}\\ \left(-\dfrac{1}{3}\right)^{50}=\left[\left(-\dfrac{1}{3}\right)^5\right]^{10}=\left(-\dfrac{1}{243}\right)^{10}\\ Vì:\left(-\dfrac{1}{125}\right)^{10}< \left(-\dfrac{1}{243}\right)^{10}\left(do:-\dfrac{1}{125}< -\dfrac{1}{243}\right)\\ =>\left(-\dfrac{1}{5}\right)^{30}< \left(-\dfrac{1}{3}\right)^{50}\left(đpcm\right)\)