Tính giá trị biểu thức
\(A=2015^{\left(1000-1^3\right).\left(1000-2^3\right)....\left(1000-20^3\right)}\)
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Tìm giá trị biểu thức
\(A=2015^{\left(1000-1^3\right).\left(1000-2^3\right)....\left(1000-20^3\right)}\)
Tính biểu thức:
\(A=\left(1-\frac{1}{1000}\right).\left(1-\frac{2}{1000}\right).\left(1-\frac{3}{1000}\right).....\left(1-\frac{50000^{1000}}{1000}\right)\)
\(F=\left(1000-1^3\right).\left(1000-2^3\right)......\left(1000-2015^3\right)\)
\(F=\left(1000-1^3\right).\left(1000-2^3\right)....\left(1000-2015^3\right)\)
\(F=\left(1000-1^3\right).\left(1000-2^3\right).....\left(1000-10^3\right)......\left(1000-2015^3\right)\)
\(F=\left(1000-1^3\right).\left(1000-2^3\right)....0.....\left(1000-2015^3\right)\)
\(F=0\)
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Tính :
\(A=\left(1000-1^3\right).\left(1000-2^3\right).\left(1000-3^3\right).........\left(1000-50^3\right)\)
Vì 103 = 1000 nên :
( 1000 - 103 ) = 0
Số nào nhân với 0 cũng bằng 0
Vậy A = 0
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B = \(\left(\frac{2017}{1000}\right)^{2017}.\left(\frac{100}{2015}\right)^{2017}:\left(—\frac{2017}{2015.10}\right)^{2017}\).Tính giá trị biểu thức
\(Tính:\left(1000-1^3\right).\left(1000-2^3\right).\left(1000-3^3\right)...\left(1000-50^3\right)\)
1.Tính C=\(\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)\left(1+\frac{1999}{3}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{3}\right)...\left(1+\frac{1000}{1999}\right)}\)
\(C=\frac{\left(1+\frac{1999}{1}\right)\left(1+\frac{1999}{2}\right)...\left(1+\frac{1999}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)...\left(1+\frac{1000}{1999}\right)}\)=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{1001.1002.1003....2999}{1.2.3...1999}}\)
=> \(C=\frac{\frac{2000.2001.2002....2999}{1.2.3...1000}}{\frac{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}}\)
=> \(C=\frac{2000.2001.2002....2999}{1.2.3...1000}.\frac{\left(1.2.3...1000\right).\left(1001.1002...1999\right)}{\left(1001.1002.1003....1999\right).\left(2000.2001.2002...2999\right)}=1\)
Đáp số: C=1
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tính G=\(\frac{\left(1+\frac{1015}{1}\right)\left(1+\frac{1015}{2}\right)\left(1+\frac{1015}{3}\right)...\left(1+\frac{1015}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)\left(1+\frac{1000}{3}\right)...\left(1+\frac{1000}{1015}\right)}\)
tinh \(G=\frac{\left(1+\frac{2015}{1}\right)+\left(1+\frac{2015}{2}\right)+...+\left(1+\frac{2015}{1000}\right)}{\left(1+\frac{1000}{1}\right)\left(1+\frac{1000}{2}\right)+....+\left(1+\frac{1000}{2015}\right)}\)