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a, 2010 x 3+ 2010 x 6 + 2010
= 2010 x ( 3 + 6 + 1)
= 2010 x 10
= 20100
b, 2011 x 89 + 10 x 2011 + 2011
= 2011 x (89 + 10 + 1)
= 2011 x 100
= 201100
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 2011 x 3+ 2011 x 6 + 2011
= 2011 x ( 3+6+1)
= 2011 x 10
= 20110
b, 2010 x 89 + 10 x 2010 + 2010
= 2010 x (89+10+1)
= 2010 x 100
= 201000
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{2009}{987654321}< \frac{2010}{987654321}\)
\(\frac{2010}{24681357}>\frac{2009}{24681357}\)
\(\Rightarrow A=B\)
\(A=\frac{2009}{987654321}+\frac{2010}{246813579}\)
\(=\frac{2009}{987654321}+\frac{2009}{246813579}+\frac{1}{246813579}\)
\(B=\frac{2009}{987654321}+\frac{1}{987654321}+\frac{2009}{246813579}\)
Có \(\frac{1}{246813579}>\frac{1}{987654321}\)
Vậy A > B
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a) Ta có : \(\frac{2010}{2011}>\frac{2010}{2011+2012}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012}\)
Nên \(\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010+2011}{2011+2012}\)=> M > N
b) P = \(\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.\left(2010+2\right)-2}{2010.2011+4020}=\frac{2011.2010+2011.2-2}{2010.2011+4020}=\)\(\frac{2011.2010+4020}{2010.2011+4020}=1\)
Nên P = 1
câu b sửa lại:\(P=\frac{2011.2012-2}{2010.2011+4020}=\frac{2011.2010+4022-2}{2010.2011+4020}=\frac{2010.2011+4020}{2010.2011+4020}=1\)
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\(M=\frac{2010}{2011}+\frac{2011}{2012}>\frac{2010}{2011+2012}+\frac{2011}{2011+2012}=\frac{2010+2011}{2011+2012}=N\)