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a) \(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}=\left(-1\right)^{3n+1}\)
b) \(B=\left(10000-1^2\right)\left(10000-2^2\right).........\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right)......\left(10000-100^2\right)....\left(10000-1000^2\right)\)
\(=\left(10000-1^2\right)\left(10000-2^2\right).....\left(10000-10000\right).....\left(10000-1000^2\right)=0\)
c) \(C=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)..........\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right).....\left(\frac{1}{125}-\frac{1}{5^3}\right)......\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)........\left(\frac{1}{125}-\frac{1}{125}\right).....\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
d) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
a.
\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)
\(=\left(-\frac{1}{2}\right)\times\left(-\frac{2}{3}\right)\times\left(-\frac{3}{4}\right)\times...\times\left(-\frac{2015}{2016}\right)\times\left(-\frac{2016}{2017}\right)\)
\(=\frac{1}{2017}\)
b.
\(\frac{2^{50}\times7^2+2^{50}\times7}{4^{26}\times112}=\frac{2^{50}\times\left(7^2+7\right)}{\left(2^2\right)^{26}\times112}=\frac{2^{50}\times\left(49+7\right)}{2^{52}\times2\times56}=\frac{56}{2^3\times56}=\frac{1}{8}\)
a. (1/2-1).(1/3-1)(1/4-1). ... .(1/2017-1)=(-1/2)(-2/3)(-3/4). ... .(-2016/2017)
Vì dãy số có 2016 số hạng âm nên tích của chúng là một số dương.
Ta có:(-1/2)(-2/3)(-3/4). ... . (-2016/2017)=1/2017
#)Giải :
a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)
b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
Ta có: \(A=\left[6.\left(-\frac{1}{3}\right)^2-3.\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)
\(=\left(6.\frac{1}{9}-\left(-1\right)+1\right):\left(\frac{-4}{3}\right)\)
\(=\left(\frac{2}{3}+2\right).\left(\frac{-3}{4}\right)\)
\(=\frac{8}{3}.\left(-\frac{3}{4}\right)\)
\(=-2\)
\(B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)
\(\Rightarrow B=\left(729-1^3\right)\left(729-3^3\right)...0...\left(729-125^3\right)\)
\(\Rightarrow B=0\)
Vì -2 < 0 nên A < B
Vậy A < B
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)
\(A=\left(-1\right)^{2n+n+n+1}\)
\(A=\left(-1\right)^{4n+1}\)
\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)
\(B=0\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=0\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)
\(D=1999^0\)
\(D=1\)
bạn ơi cho mk hỏi 1 bài làm giúp mk đc ko vậy ạ
2n là số chẳn , n và n+1 n chẳn thì n+1 là lẻ và ngược lại nên A = -1