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a) \(9\cdot2^3-72:2^2\)
\(=72-18=54\)
b) \(125\cdot26+5^3\cdot32+25\cdot42\cdot5\)
\(=125\cdot26+125\cdot32+125\cdot42\)
\(=125\cdot\left(26+32+42\right)=125\cdot100=12500\)
c) \(5871:\left[928-\left(247-82\right)\cdot5\right]+2012^0\)
\(=5871:\left[928-\left(247-82\right)\cdot5\right]+1\)
\(=5871:\left(928-165\cdot5\right)+1\)
\(=5871:103+1\)
\(=58\)
d) \(2012+78+\left(-178\right)+\left(-2012\right)\)
\(=\left[2012+\left(-2012\right)\right]+\left[78+\left(-178\right)\right]\)
\(=-100\)
a, \(9.2^3-\frac{72}{2^2}=9.8-\frac{72}{4}=72-18=54\)
b, \(125.26+5^3.32+25.42.5\)
\(=125.26+125.32+125.42\)
\(=125.\left(26+32+42\right)\)
\(=125.100\)
\(=12500\)
c, \(\frac{5871}{\left[928-\left(247-82\right).5\right]}+2012^0\)
\(=\frac{5871}{103}+1\)
\(=57+1\)
\(=58\)
d, \(2012+78+\left(-178\right)+\left(-2012\right)\)
\(=\left[2012+\left(-2012\right)\right]+\left[78+\left(-178\right)\right]\)
\(=0+100\)
\(=100\)
b/1.2.3.....2013 - 1.2.3....2012-1.2.3...20122
= (1.2.3...2012.2013-1.2.3...2012)-1.2.3...2012.2012
= 1.2.3...2012(2013-1-2012)
=1.2.3...2012.0=0
Câu 2:
a: \(=9\cdot8-72:4=72-18=54\)
b: \(=125\cdot26+125\cdot32+125\cdot42\)
\(=125\left(26+32+42\right)=125\cdot100=12500\)
c: \(=5871:\left(928-825\right)+1=5871:103+1=57+1=58\)
d: \(=2012+78-178-2012=-100\)