Вычислите; 1) (55/57-51/95):(-122/133); 2) (-60/259):(-45/148+75/222); 3)...

Решите задачу:

$( \frac{55}{57} - \frac{51}{95} ) :( -\frac{122}{133} )= \frac{55*5 - 51*3}{285} : (- \frac{122}{133} )= \\ = \frac{275-153}{285} * ( - \frac{133}{122}) = -\frac{122*133}{285*122} = -\frac{1*7}{15*1} = -\frac{7}{15}$

$(- \frac{60}{259} ): (- \frac{45}{148} + \frac{75}{222} ) = (- \frac{60}{259} ):( \frac{75*2 - 45*3}{444} ) = (- \frac{60}{259} ) : \frac{15}{444} = \\ \\ =- \frac{60*444}{259*15} = -\frac{4*12}{7*1} =- \frac{48}{7} =-6 \frac{6}{7}$

$(\frac{8}{225} - \frac{4}{105} ) : ( \frac{6}{35} - \frac{2}{45} )= \frac{8*7-4*15}{1575} : \frac{6*9-7*7}{315} = \\ \\ =- \frac{4}{1575} : \frac{5}{315} =- \frac{4*315}{1575*5} = -\frac{4*1}{5*5} = -\frac{4}{25} =-0,16$

$( \frac{4}{51} - \frac{1}{136} ) : (- \frac{5}{34} - \frac{7}{51} )= \frac{4*8-1*3}{408} : \frac{- 5*3 - 7*2}{102} = \\ \\ = - \frac{29*102}{408*29} = - \frac{1}{4} =-0,25$