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30 tháng 11 2017
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0......\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=0\)
19 tháng 10 2019
\(=\)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{5^3}\right)\)\(...\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(\left(\frac{1}{125}-\frac{1}{1^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{2^3}\right)\) \(.\) \(\left(\frac{1}{125}-\frac{1}{3^3}\right)\) \(.\) \(0\) \(....\) \(\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\) \(0\)
1 tháng 8 2019
\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)\)
\(=0\)
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) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
VN
Võ Ngọc Phương
VIP
1 tháng 10 2023
Đây nha:
\(\left(x-\dfrac{1}{8}\right)^3=-\dfrac{8}{125}\)
\(\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow x-\dfrac{1}{8}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}+\dfrac{1}{8}=\dfrac{-16}{40}+\dfrac{5}{40}\)
\(x=\dfrac{-11}{40}\)
\(#WendyDang\)
VN
Võ Ngọc Phương
VIP
1 tháng 10 2023
\(\left(x-\dfrac{1}{3}\right)^3=\dfrac{-8}{125}\)
\(\left(x-\dfrac{1}{3}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}+\dfrac{1}{3}=\dfrac{-6}{15}+\dfrac{5}{15}\)
\(x=\dfrac{-1}{15}\)
T
13 tháng 7 2019
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
21 tháng 6 2017
b) ( 2/5 )^ x-1 : ( 2/5 )^2 = ( 2/5 )^3
(2/5)^ x-1 = ( 2/5) ^3 . (2/5)^2
(2/5)^ x-1 = ( 2/5 ) ^ 6
=> x-1 = 6
=> x= 7
Bài làm:
( x + 1) ^3 = -125
<=> ( x + 1)^3= (-5)^3
<=> x + 1 = -5
<=> x = -5 -1
<=> x= -6
# Chúc bạn hok tốt nha
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