a) \(\dfrac{43}{47}=\dfrac{47-4}{47}=\dfrac{47}{47}-\dfrac{4}{47}=1-\dfrac{4}{47}\)

\(\dfrac{53}{57}=\dfrac{57-4}{57}=\dfrac{57}{57}-\dfrac{4}{57}=1-\dfrac{4}{57}\)

Vì \(\dfrac{4}{47}>\dfrac{4}{57}\)=>\(1-\dfrac{4}{47}< 1-\dfrac{4}{57}\)

=>\(\dfrac{43}{47}< \dfrac{53}{57}\)

b) Ta chọn phân số trung gian là: \(\dfrac{64}{81}\)

Vì \(\dfrac{64}{81}>\dfrac{64}{85}\)

Mà \(\dfrac{64}{81}< \dfrac{73}{81}\)

=> \(\dfrac{64}{85}< \dfrac{64}{81}< \dfrac{73}{81}\)

Vậy \(\dfrac{64}{85}< \dfrac{73}{81}\)

c) Ta chọn phân số trung gian là: \(\dfrac{18}{37}\)

Vì \(\dfrac{18}{37}>\dfrac{15}{37}\)

Mà \(\dfrac{18}{37}< \dfrac{18}{31}\)

=> \(\dfrac{15}{37}< \dfrac{18}{37}< \dfrac{18}{31}\)

Vậy \(\dfrac{18}{31}>\dfrac{15}{37}\)

Phần d) mình không biết lời giải

Mình chỉ biết \(\dfrac{12}{47}>\dfrac{19}{77}\) thôi !

a) Ta có:

\(\dfrac{2012}{2013}+\dfrac{1}{2013}=1\)

\(\dfrac{2013}{2014}+\dfrac{1}{2014}=1\)

Vì \(\dfrac{1}{2013}>\dfrac{1}{2014}\) nên \(\dfrac{2012}{2013}< \dfrac{2013}{2014}\)

b) Ta có:

\(\dfrac{1006}{1007}+\dfrac{1}{1007}=1\)

\(\dfrac{2013}{2015}+\dfrac{2}{2015}=1\)

Vì \(\dfrac{1}{1007}=\dfrac{2}{2014}>\dfrac{2}{2015}\) 

nên \(\dfrac{1006}{1007}< \dfrac{2013}{2015}\)

c) ta có:

\(1-\dfrac{64}{73}=\dfrac{9}{73}=\dfrac{153}{1241}\)

\(1-\dfrac{45}{51}=\dfrac{2}{17}=\dfrac{146}{1241}\)

Vì \(\dfrac{153}{1241}>\dfrac{146}{1241}\) nên \(\dfrac{63}{73}>\dfrac{45}{51}\)

a) 2012/2013 và 2013/2014

1-2012/2013=1/2013

1-2013/2014=1/2014

Vì 1/2013> 1/2014 nên 2012/2013<2013/2014

b) 1006/1007 và 2013/2015

1-1006/1007=1/1007=2/2014

1-2013/2015=2/2015

Vì 2/2014>2/2015 nên 1006/1007<2013/2015

c) 64/73 và 45/51

1-64/73=9/73=18/146

1-45/51=2/17=18/153

Vì 18/146> 18/153 nên 64/73<45/51

a) Ta có: \(\frac{13}{57}=1-\frac{44}{57}\)

              \(\frac{29}{73}=1-\frac{44}{73}\)

Ta thấy:    \(\frac{44}{57}>\frac{44}{73}\)\(\Rightarrow\)\(1-\frac{44}{57}< 1-\frac{44}{73}\)

Vậy   \(\frac{13}{57}< \frac{29}{73}\)

Bài 1:

\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)

\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)

Bài 2:

\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)

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