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NN
21 tháng 8 2020
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) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
24 tháng 7 2017
\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)
\(A=\left(-1\right)^{2n+n+n+1}\)
\(A=\left(-1\right)^{4n+1}\)
\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)
\(B=0\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=0\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)
\(D=1999^0\)
\(D=1\)
NT
0
NN
3 tháng 7 2019
=-(0,1+0,01+0,001+0,0001+0,00001)
=-0,11111
Đây là cách đơn giản nhất
NT
3
HT
29 tháng 9 2015
B = \(-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-...-\frac{1}{1000000}\)
B = \(-\left(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^6}\right)\)
Đặt A = \(\frac{1}{10}+\frac{1}{10^2}+\frac{1}{10^3}+...+\frac{1}{10^6}\)
10A = \(1+\frac{1}{10}+\frac{1}{10^2}+...+\frac{1}{10^5}\)
9A = 10A - A = \(1-\frac{1}{10^6}\)
=> A = \(\frac{1-\frac{1}{10^6}}{9}\)
=> B = \(-\left(\frac{1-\frac{1}{10^6}}{9}\right)\)
17 tháng 10 2018
C=(0,1+0,01+0,001+...+0,000001)=-0,111111
mình ko chép đề bài
19 tháng 9 2018
\(C=-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}-\frac{1}{100000}\)
\(10C=-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\)
\(10C-C=\left(-1-\frac{1}{10}-\frac{1}{100}-\frac{1}{1000}-\frac{1}{10000}\right)-\left(\frac{-1}{10}-\frac{1}{100}-...-\frac{1}{100000}\right)\)
\(9C=-1+\frac{1}{100000}\)
\(C=\frac{\frac{1}{100000}-1}{9}\)
10000-100^2 =10000-10000=0
Vậy tích trên bằng 0
(10000-100^2)=0 => h = 0