Question

resolver (con procedimiento) el ">" sig elevado


. 33/7 + 45/8 x (5+8-2x2) 2>3 + 7 % 7


.4x6/2+7+1/3-5/4



.35-5+19x2%4+54-12


.1/2 - 3/4 + 1/3


.1/7 + 2/4 - 5/2


5/3 - 4/3 -35/3

Answer 1

$\frac{33}{7}+\frac{45}{8}*(5+8-2*2)2>3+\frac{7}{7}\\\frac{33}{7}+\frac{45}{8}*(5+8-4)2>3+1\\\frac{33}{7}+\frac{45}{8}*(5+4)2>4\\\frac{33}{7}+\frac{45}{8}*(9)2>4\\\frac{579}{56}*18>4\\\frac{5211}{28}>4$

$4*\frac{6}{2}+7+\frac{1}{3}-\frac{5}{4}\\4*3+7-\frac{11}{12}\\12+\frac{73}{12}=\frac{217}{12}$

$35-5+19*\frac{2}{4}+54-12\\30+19*\frac{1}{2}+42\\\frac{19}{2}+72=\frac{163}{2}$

$\frac{1}{2}-\frac{3}{4}+\frac{1}{3}=-\frac{1}{4}+\frac{1}{3}=\frac{1}{12}$

$\frac{1}{7}+\frac{2}{4}-\frac{5}{2}=\frac{1}{7}+\frac{1}{2}-\frac{5}{2}=\frac{9}{14}-\frac{5}{2}=-\frac{13}{7}$

$\frac{5}{3}-\frac{4}{3}-\frac{35}{3}=\frac{5-4-35}{3}=-\frac{34}{3}$