Tính rồi so sánh \(A\) và \(B\) biết rằng

\(A = \dfrac{{1313}}{{1818}} + \dfrac{{19}}{{12}} \times 1\dfrac{3}{5} \times 1\dfrac{7}{{38}};\)

\(B = \,\dfrac{{2121}}{{2323}}:\dfrac{{49}}{{46}} + \dfrac{{24}}{{35}} \times \dfrac{5}{{16}}.\)

Ta có

$A = \dfrac{{1313}}{{1818}} + \dfrac{{19}}{{12}} \times 1\dfrac{3}{5} \times 1\dfrac{7}{{38}}$

$A = \dfrac{{1313}}{{1818}} + \dfrac{{19}}{{12}} \times \dfrac{8}{5} \times \dfrac{{45}}{{38}}$

$A = \dfrac{{13 \times 101}}{{18 \times 101}} + \,\dfrac{{{{{19}}} \times {{8}} \times {{{45}}}}}{{{{{12}}} \times {{5}} \times {{{38}}}}}$

$A = \dfrac{{13 \times { {{101}}}}}{{18 \times { {{101}}}}} + \,\dfrac{{1 \times {{2}} \times {{9}}}}{{{ {3}} \times 1 \times { {2}}}}$

$A = \dfrac{{13}}{{18}} + \,3$

$A = 3\dfrac{{13}}{{18}}$

Lại có 

$B = \,\dfrac{{2121}}{{2323}}:\dfrac{{49}}{{46}} + \dfrac{{24}}{{35}} \times \dfrac{5}{{16}}$

$B = \,\dfrac{{2121}}{{2323}} \times \dfrac{{46}}{{49}} + \dfrac{{24}}{{35}} \times \dfrac{5}{{16}}$

$B = \,\dfrac{{21 \times 101}}{{23 \times 101}} \times \dfrac{{46}}{{49}} + \dfrac{{{{{24}}} \times {{5}}}}{{{{{35}}} \times { {{16}}}}}$

$B = \,\dfrac{{21 \times { {{101}}}}}{{23 \times { {{101}}}}} \times \dfrac{{46}}{{49}} + \dfrac{3}{{14}}$

$B = \,\dfrac{{21}}{{23}} \times \dfrac{{46}}{{49}} + \dfrac{3}{{14}}$

$B = \,\dfrac{ {{21} \times {{{46}}}}}{{{{{23}}} \times {{{49}}}}} + \dfrac{3}{{14}}$

$B = \dfrac{6}{7} + \dfrac{3}{{14}}$

$B = \dfrac{{12}}{{14}} + \dfrac{3}{{14}}$$= \dfrac{{15}}{{14}} $$= 1\dfrac{1}{{14}}$

Hỗn số \(3\dfrac{{13}}{{18}}\) có  phần nguyên là \(3\) và hỗn số \(1\dfrac{1}{{14}}\) có  phần nguyên là \(1\).

Vì \(3 > 1\) nên \(3\dfrac{{13}}{{18}} > 1\dfrac{1}{{14}}\).

Vậy \(A > B\).