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1+2+3+4+5+6+7+8+9+10=55
11+12+13+14+15+16+17+18+19+20=155
1+2+3+4+5+6+7+8+9+10+11+12+13+14 +15+16+17+18+19+20+21+22+23+24+25+26+27+28+29+30-50-53=362
\(1+2+3+4+5+6+7+8+9+10=55\)
\(11+12+13+14+15+16+17+18+19+20=155\)
12) 17 – 11 – 14 – (-39)
=17-11-14+39
=6-14+39
=-8+39
=31
13) −127 - 18.( 5 - 6)
=-127-18.1
=-145
14) 100 + (+430) + 2145 + (-530)
=530+2145-530
=(530-530)+2145
=0+2145
=2145
15) (+12).13 + 13.(-22)
=13.[12+(-22)]
=13.(-10)
=-130
16) {[14 : (-2)] + 7} : 2012
={-7+7}:2012
=0:2012
=0
17) 13 – 18 – (- 42) - 15
=13-18+42-15
=-5+42-15
=37-15
=22
18) 369 – 4[(-5) + 4.(-8)]
=369-4[-5+(-32)]
=369-4(-37)
=369+148
=517
19) (-12).(-13) +13.(-29)
=12.13+13.(-29)=13.[12+(-29)]=13.(-17)=-221
20) 125 – 4[ 3 – 7.(-2) ]
=125-4[3+14]
=125-4.17
=125-68
=57
21) (-14).9 – 13.(-9)
=(-14).9+13.9
=9.[-14+13]
=9.(-1)
=-9
3: \(=\dfrac{13}{5}\left(-\dfrac{3}{14}+\dfrac{2}{5}+\dfrac{-11}{14}+\dfrac{3}{5}\right)\)
=0
\(A=\dfrac{1+2+3+...+19}{11+12+13+...+29}=\dfrac{\dfrac{\left(1+19\right)\left(19-1+1\right)}{2}}{\dfrac{\left(11+29\right)\left(29-11+1\right)}{2}}=\dfrac{190}{380}=\dfrac{1}{2}\)
Vậy \(A=\dfrac{1}{2}\)