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Ta có: \(B=\frac{3.5.7.11.13.17-10101}{1313130+20202}\)
\(\Rightarrow B=\frac{255255-10101}{1313130+20202}\)
\(\Rightarrow B=\frac{1001.255-10101}{10101.13.10+10101.2}\)
\(\Rightarrow B=\frac{1001.255-10101}{10101.\left(13.10+2\right)}\)
\(\Rightarrow B=\frac{1001.255-10101}{10101.132}\)
\(\Rightarrow B=\frac{10101.255-19201}{10101.132}\)
\(10101.\left(\frac{5}{10101}-\frac{5}{20202}+\frac{5}{30303}+\frac{5}{40404}\right)\)
\(=10101.\frac{5}{10101}-10101.\frac{5}{20202}+10101.\frac{5}{30303}+10101.\frac{5}{40404}\)
\(=5-\frac{5}{2}+\frac{5}{3}+\frac{5}{4}\)
\(=\frac{60}{12}-\frac{30}{12}+\frac{20}{12}+\frac{15}{12}\)
\(=\frac{60-30+20+15}{12}\)
\(=\frac{65}{12}\)
Bài 1: Rút gọn phân số:
A=\(\frac{11}{21.10^2+10^2}\)
B=\(\frac{3.5.7.11.13.37-10101}{1313130+20202}\)
\(A=\frac{11}{21.10^2+10^2}=\frac{11}{10^2.\left(21+1\right)}=\frac{11}{10^2.22}=\frac{11}{10^2.11.2}=\frac{1}{10^2.2}\)
\(B=\frac{3.5.7.11.13.37-10101}{1313130+20202}=\frac{\left(3.5.7.11.13.37\right).5-10101.1}{130.10101+10101.2}=\frac{10101.\left(5-1\right)}{10101.\left(130+2\right)}=\frac{4}{132}=\frac{1}{33}\)
(14568+3579)-(14563+3564)
= 14568+3579-14563-3564
=(14568-14563)+(3579-3564)
=5+15=20
\(=20.\left(-9\right)+\left(-20\right).21\)
\(=-180+\left(-420\right)\)
\(=-600\)
a) =20.(-9)+(-20).21
=20.(-9)+20.(-21)
=20[(-9)+(-21)]
=20.(-30)
=-600
a) 544544 – 444444 = 544.1001 – 444.1001 = 1001.(544 – 444) = 1001.100 = 100100
b) 131313 – 10101 – 20202 = 10101.13 – 10101 – 10101.2
= 10101.(13 – 1 – 2) = 10101.10 = 101010