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b) Ta có: \(A=\dfrac{1012+1}{1013+1}\)
\(\Leftrightarrow A-1=\dfrac{1012+1-1013-1}{1013+1}\)
\(\Leftrightarrow A-1=\dfrac{-1}{1013+1}\)
Ta có: \(B=\dfrac{1011+1}{1012+1}\)
\(\Leftrightarrow B-1=\dfrac{1011+1-1012-1}{1012+1}\)
\(\Leftrightarrow B-1=\dfrac{-1}{1012+1}\)
Ta có: \(1013+1>1012+1\)
\(\Leftrightarrow\dfrac{1}{1013+1}< \dfrac{1}{1012+1}\)
\(\Leftrightarrow\dfrac{-1}{1013+1}>\dfrac{-1}{1012+1}\)
\(\Leftrightarrow A-1>B-1\)
hay A>B
Vậy: A>B
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\(A=\frac{595959}{454545}-\frac{14141414}{15151515}\)
\(A=\frac{595959:10101}{454545:10101}-\frac{14141414:1010101}{15151515:1010101}\)
\(A=\frac{59}{45}-\frac{14}{15}\)
\(A=\frac{59}{45}-\frac{42}{45}=\frac{17}{45}\)
\(B=\frac{65+891+135+909}{731-47+69-253}\)
\(B=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)
\(B=\frac{200+1800}{800-300}=\frac{2000}{500}=4\)
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a)387+(-224)+(-87)=[387+(-87)]+(-224)=300+(-87)=213
b)(-75)+329+(-25)=[(-75)+(-25)]+329=-100+329=229
e)251+(-114)+(-151)+(-216)=[251+(-151)]+[(-114)+(-216)]=100+(-330)=-230
j)(-137)+77+(-253)+37-47=[(-137)+37]+77+[(-253)-47]=-100+77-300=177-300=-123
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a) 951 . 23 = 21 873
b) 47 . 273 = 12 831
c) 845 . 253 = 213 785
d) 1 356 . 125 = 169 500
2013 x ( 1013 - 13 ) + 100 x ( 253 + 47 )
= 2013 x 1000 + 100 x 300
= 2 013 000 + 3 000
= 2 016 000
2013 x ( 1013 - 13 ) + 100 x ( 253 + 47 )
2013 x 1000 + 100 x 300
2013000 + 300000
2313000