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1: \(-87+\left(-12\right)-\left(-487\right)+512\)
\(=-87+487+512-12\)
\(=\left(-87+487\right)+\left(512-12\right)\)
=400+500
=900
2: \(942-2567+2563-1942\)
\(=\left(942-1942\right)-\left(2567-2563\right)\)
=-1000-4
=-1004
3: \(1152-\left(374-1152\right)+\left(-65+374\right)\)
\(=1152-374+1152-65+374\)
\(=\left(1152+1152\right)+\left(374-374\right)-65\)
=2304-65
=2239
1: −87+(−12)−(−487)+512−87+(−12)−(−487)+512
=−87+487+512−12=−87+487+512−12
=(−87+487)+(512−12)=(−87+487)+(512−12)
=400+500
=900
2: 942−2567+2563−1942942−2567+2563−1942
=(942−1942)−(2567−2563)=(942−1942)−(2567−2563)
=-1000-4
=-1004
3: 1152−(374−1152)+(−65+374)1152−(374−1152)+(−65+374)
=1152−374+1152−65+374=1152−374+1152−65+374
= (1152+1152)+(374−374)−65=(1152+1152)+(374−374)−65
= 2304-65
= 2239
A=1-1/(2013*2014)
B=1-1/(2014*2015)
2013*2014<2014*2015
=>1/2013*2014>1/2014*2015
=>-1/2013*2014<-1/2014*2015
=>A<B
A=2011^2012-2011^2011= 2011^2011 * 2011 -2011^2011= 2011^2011 *(2011-1)= 2011^2011 *2010
B=2011^2013-2011^2012=2011^2012*2011- 2011^2012= 2011^2012 *(2011-1) = 2011^2012 *2010
vì 2011^2011*2010 < 2011^2012*2010 nên A<B
Ta có : 2011^2013 x M = (2010^2012 x 2011 + 2011^2013)^2013 > (2010^2013 + 2011^2013)^2013 = N x (2010^2013 + 2011^2013)
Do đó: 2011^2013 x M > N x (2010^2013 + 2011^2013)
<=> M > N x [(2010/2011)^2013 + 1] ==> M > N (điều phải chứng minh)
\(A>\dfrac{2^{2018}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{3^{2019}}{2^{2018}+3^{2019}+5^{2020}}+\dfrac{5^{2020}}{5^{2020}+2^{2018}+3^{2019}}=1\)
\(B< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2019\cdot2020}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)
=>B<1
=>A>B