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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.