Tính: \(\left(\frac{3}{2}\right)^{2012}:\left(\frac{9}{25}\right)^{1000}\)là ?
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\(\left(\frac{3}{5}\right)^{2012}:\left(\frac{9}{25}\right)^{1000}=\left(\frac{3}{5}\right)^{2012}:\left(\frac{3}{5}\right)^{2000}=\left(\frac{3}{5}\right)^{12}\)
\(\left(\frac{3}{5}\right)^{2003}:\left(\frac{9}{25}\right)^{1000}\)
\(=\left(\frac{3}{5}\right)^{2003}:\left(\left(\frac{3}{5}\right)^2\right)^{1000}\)
\(=\left(\frac{3}{5}\right)^{2003}:\left(\frac{3}{5}\right)^{2000}\)
\(=\left(\frac{3}{5}\right)^3\)
\(=\frac{27}{125}\)
\(\left(\frac{3}{5}\right)^{2003}:\left(\frac{9}{25}\right)^{1000}\)
\(=\frac{3}{5}.\left[\left(\frac{3}{5}\right)^2\right]^{1000}:\left(\frac{9}{25}\right)^{1000}\)
\(=\frac{3}{5}.\left(\frac{9}{25}\right)^{1000}:\left(\frac{9}{25}\right)^{1000}\)
\(=\frac{3}{5}\)
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) \(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-10^3\right)}\)
\(=1999^{\left(1000-1^3\right)\left(1000-2^3\right)........\left(1000-1000\right)}=1999^0=1\)
Đặt A=4+6+8+...+2012
Số số hạng của dãy là: (2012-4)\(\div\)2+1=1005
Tổng A=(2012+4)\(\times\)1005\(\div\)2=1013040
\(\Rightarrow\)1013040\(\times\frac{1}{1000}\times\left(\frac{1}{2}+\frac{3}{4}+\frac{5}{6}\right)=\) 1013040\(\times\frac{1}{1000}\times\frac{25}{12}=\)\(\frac{4221}{2}\)=2110,5
\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)
=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)
=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)
Đáp số: C=1