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A

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

$a)\;\frac{{1995}}{{1996}}\;\,...\,\;\frac{{1996}}{{1997}}$

Ta có: $1 - \frac{{1995}}{{1996}} = \frac{1}{{1996}};\;\;1 - \frac{{1996}}{{1997}} = \frac{1}{{1997}}.$

Vì $\frac{1}{{1996}} > \frac{1}{{1997}}$ nên $\frac{{1995}}{{1996}} < \frac{{1996}}{{1997}}.$

$c)\;\;\frac{{2121}}{{3737}}\;\,...\,\;\frac{{212121}}{{373737}}$

Ta có: $\frac{{2121}}{{3737}} = \frac{{2121:101}}{{3737:101}} = \frac{{21}}{{37}}$

$\;\frac{{212121}}{{373737}} = \frac{{212121:10101}}{{373737:10101}} = \frac{{21}}{{37}}$

Vậy $\frac{{2121}}{{3737}} = \frac{{212121}}{{373737}}\;\left( { = \frac{{21}}{{37}}} \right).$

$e)\,\,\frac{{18}}{{17}}\,\,...\,\,\frac{{13}}{{27}}$   

Vì $\frac{{18}}{{17}} > 1,\;\;\frac{{13}}{{27}} < 1$ nên $\frac{{18}}{{17}} > \frac{{13}}{{27}}.$

$b)\;\frac{{16}}{{27}}\;\,...\,\;\frac{{15}}{{29}}$

Vì $\frac{{16}}{{27}} > \frac{{16}}{{29}},\;\;\frac{{15}}{{29}} < \frac{{16}}{{29}}$ nên $\frac{{16}}{{27}} > \frac{{15}}{{29}}.$

$d)\;\frac{{2010}}{{2011}}\;\,...\,\;\frac{{17}}{{16}}$

Vì $\frac{{2010}}{{2011}} < 1,\;\;\frac{{17}}{{16}} > 1$

Nên $\frac{{2010}}{{2011}} < \frac{{17}}{{16}}.$

$f)\;\frac{{327}}{{326}}\;\,...\,\;\frac{{326}}{{325}}$

Ta có:

$\frac{{327}}{{326}} - 1 = \frac{1}{{326}},\;\;\frac{{326}}{{325}} - 1 = \frac{1}{{325}}.$

Vì: $\frac{1}{{326}} < \frac{1}{{325}}$ nên $\frac{{327}}{{326}} < \frac{{326}}{{325}}.$

Chọn A.