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Ta có:
\(\left(\frac{a}{b}\right)^3=\frac{1}{1000}=\left(\frac{1}{10}\right)^3\)
\(\Rightarrow\frac{a}{b}=\frac{1}{10}\Rightarrow\frac{a}{1}=\frac{b}{10}\)
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a}{1}=\frac{b}{10}=\frac{b-a}{10-1}=\frac{36}{9}=4\)
\(\Rightarrow\begin{cases}a=4.1=4\\b=4.10=40\end{cases}\)
Vậy a = 4; b = 10
Ta có : \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{20}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{19}{20}\)
\(=\frac{1.2.3.....19}{2.3.4.....20}\)
\(=\frac{1}{20}>\frac{1}{21}\)
Có : (a/b)^3 = 1/1000 =(1/10)^3
<=> a/b = 1/10
<=> a = b/10
Khi đó : b - b/10 = 36
<=> 9/10 . b = 36
<=> b = 36 : 9/10 = 40
<=> a = b/10 = 40/10 = 4
Vậy a= 4; b= 40
(\(\frac{a}{b}\))3=\(\frac{1}{1000}\)=(\(\frac{1}{10}\))3 => a/b=1/10 hay b=10a
=> 10a-9a=36 <=> 9a=36 => a=4; b=36+4=40
ĐS: a=4; b=40
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) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(\left(\frac{a}{b}\right)^3=\left(\frac{1}{10}\right)^3\)
=> \(\frac{a}{b}=\frac{1}{10}\) => \(\frac{a}{1}=\frac{b}{10}=\frac{b-a}{10-1}=\frac{36}{9}=4\)
=> a = 4; b = 4.10=40