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\(A=\frac{65+891+135+909}{731-47+69-253}\)
\(A=\frac{65+135+891+909}{731+69-47-253}\)
\(A=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)
\(A=\frac{200+1800}{800-300}\)
\(A=\frac{2000}{500}=4\)
A = \(\frac{65+891+135+909}{731-47+69-253}\)
A = \(\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)
A = \(\frac{200+1800}{800-300}\)
A = \(\frac{2000}{500}\)
A = 4
\(C=\frac{65+891+135+909}{731-47+69-253}\)
\(C=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)
\(C=\frac{200+1800}{800-300}\)
\(C=\frac{2000}{500}=4\)
a)2006x125+1000=250,75x125x8+1000=250,75x1000+1000=1000(250,75+1)=1000x251,75=251750
731-47+69-253=(731+69)-(47+253)=800-300=500
b)3,54x73+0,46x25+3,54x26+0,46x75+3,54
=3,54(73+26+1)+0,46(25+75)
=3,54x100+0,46x100
=354+46=400
#)Giải :
A, \(\frac{254x399-145}{254+399x253}\)
\(=\frac{253x399+399-145}{254+399x253}\)
\(=\frac{253x399+254}{254+399x253}\)
\(=1\)
B, \(\frac{5932+6001x5931}{5931x6001-69}\)
\(=\frac{5932+6001x5931}{\left(5931+1\right)x6001-69}\)
\(=\frac{5932+6001x5931}{5931x6001+6001-69}\)
\(=\frac{5932+6001x5932}{5932x6001+5932}\)
\(=1\)
#~Will~be~Pens~#
\(\frac{254x399-145}{254+399x253}=\frac{\left(253+1\right)x399-145}{254+399x253}=\frac{253x399+1x399-145}{254+399x253}=\frac{253x399+254}{254+399x253}\)
\(=1\)
a) Ta có\(\frac{47}{15}=3\left(\frac{2}{15}\right)\) và \(\frac{65}{21}=3\left(\frac{2}{21}\right)\)
Vì \(\frac{2}{15}>\frac{2}{21}\) nên \(3\left(\frac{2}{15}\right)>2\left(\frac{2}{21}\right)\) hay\(\frac{47}{15}>\frac{65}{21}\)
Vậy \(\frac{47}{15}>\frac{65}{21}\)