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\(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-10^3\right)..\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...\left(1000-1000\right)..\left(1000-15^3\right)}\)
= \(2009^{\left(1000-1^3\right).\left(1000-2^3\right)...0..\left(1000-15^3\right)}\)
= \(2009^0\)
= \(1\)
\(\frac{1}{2^2}-1=\frac{1-2^2}{2^2}=\frac{\left(1-2\right)\left(1+2\right)}{2^2}=-1.\frac{3}{2^2}\)
\(\frac{1}{3^2}-1=\frac{1-3^2}{3^2}=\frac{\left(1-3\right)\left(1+3\right)}{3^2}=-2.\frac{4}{3^2}\)
Đặt nguyên biểu thức là B , ta có :
\(B=\left[-1.\left(-2\right).\left(-3\right)...\left(-99\right)\right].\frac{3.4.5...101}{\left(2.3.4.5...100\right)^2}\)
\(B=-\left(1.2.3...99\right).\frac{3.4.5...101}{\left(2.3.4.5...100\right)^2}\)
B=\(\frac{-2.\left(3.4.5...99\right)^2.100.101}{2^2\left(3.4.5...99\right)^2.100^2}=\frac{-101}{200}\)
Ta thấy trong tích có thừa số ( 216 - 6\(^3\)) = 216 - 216 = 0 nên tích bằng 0
Vậy M = 0 + ( 216 x 64 + 216 x 36 )
M = 216 . ( 64 + 36 )
M = 216 x 100
M = 21600
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(A=\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right).....\left(\frac{1}{100^2}-1\right)\)
=> \(-A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{100^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.....\frac{100^2-1}{100^2}\)
\(=\frac{1.3}{2^2}.\frac{2.4}{3^2}.....\frac{99.101}{100^2}\)
\(=\frac{1.2....99}{2.3....100}.\frac{3.4....101}{2.3....100}\)
\(=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
=> \(A=-\frac{101}{200}< -\frac{1}{2}\)
A=[(1+2+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x (2,4x42 - 21x4,8)] / 1+1/2+1/3+...+1/100
= [(1+2+3+...+100) x (1/2 - 1/3 - 1/4-1/5) x (2,4x2x21 - 21x2x 4,8)] / 1+1/2+1/3+...+1/100
=[(1+2+3+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x 0] / 1+1/2+1/3+...+1/100
=0 / 1+1/2+1/3+...+1/100 = 0
\(=2^{\left(100-1^2\right)\left(100-2^2\right)...\left(100-10^2\right)...\left(100-15^2\right)}\)
=20=1