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Ta có
2+4-6-8+10+12-14-16+18+20-22-24+...-2008
=(2+4-6-8)+(10+12-14-16)+(18+20-22-24)+...+(2002+2004-2006-2008)
=(-8)+(-8)+(-8)+...+(-8) (251 số hạng (-8) )
=(-8).251
=-2008
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\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
= \(18^8-18^2+1=1\)
\(9^8.2^8-\left(18^4-1\right)\left(18^4+1\right)\)
\(=18^8-\left(18^8-1\right)\)
\(=18^8-18^8+1\)
\(=1\)
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a) 1632+74x163+372 = 163(163+74)+ 372
= 163.237 + 1369
= 40000
b)98x28-(184-1)(184+1) = (9.2)8 - 188 + 12
= 188 - 188 + 1
= 1
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\(\frac{\left(-2\right)^{18}.8^{15}.16^4}{32^{16}}\)
\(=\frac{2^{18}.2^{45}.2^{16}}{2^{80}}\)
\(=\frac{2^{79}}{2^{80}}=\frac{1}{2}\)
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A/5+10+15+...+1500
=5+10+15+...+1500 ta có:1500-5:5+1=300(số hạng)
=(5+1500)x300:2=225750
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Dãy trên có 108 số số hạng
C = (2+4-6-8)+(10+12-14-16)+........+(210+212-214-216)
C= (-8)+(-8)+........+(-8)
C= (-8) x 108
C= (-864)
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Đặt A = \(\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\)...\(-\frac{1}{1024}\)
A= \(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)....\(-\frac{1}{2^{10}}\)
2A=\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^9}\)
2A-A=(\(\frac{1}{1}\)\(-\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\)...\(-\frac{1}{2^{10}}\)) \(-\)(\(\frac{1}{2^1}-\frac{1}{2^2}-\frac{1}{2^3}-\frac{1}{2^4}-\)..\(-\frac{1}{2^9}\))
A=\(1+\frac{1}{2^{10}}\)
A= \(\frac{1025}{1024}\)