Question

26 ,если что,решать по действиям 2 последних примера(которые длинные,в самом низу)

Answer 1

а)

$20\frac56-\left(1\frac45\cdot\left(10\frac7{12}-7\frac{19}{36}\right)-\frac{17}{56}\cdot7\frac{11}{17}\right)\cdot1\frac25=\boxed{16\frac{23}{60}}\\\\1)\;10\frac7{12}-7\frac{19}{36}=\frac{127}{12}-\frac{271}{36}=\frac{381}{36}-\frac{271}{36}=\frac{110}{36}=\frac{55}{18};\\\\2)\;1\frac45\cdot\frac{55}{18}=\frac95\cdot\frac{55}{18}=\frac{11}2;\\\\3)\;\frac{17}{56}\cdot7\frac{11}{17}=\frac{17}{56}\cdot\frac{130}{17}=\frac{65}{28};\\\\4)\;\frac{11}2-\frac{65}{28}=\frac{154}{28}-\frac{65}{28}=\frac{89}{28};$

$5)\frac{89}{28}\cdot1\frac25=\frac{89}{28}\cdot\frac75=\frac{89}4\cdot\frac15=\frac{89}{20};\\\\6)\;20\frac56-\frac{89}{20}=\frac{125}6-\frac{89}{20}=\frac{1250}{60}-\frac{267}{60}=\frac{983}{60}=16\frac{23}{60}$

б)

$\left(\left(3\frac13+5\frac79\right)\cdot2\frac14-1\frac1{15}\cdot\left(3\frac56-\frac7{12}+9\frac78\right)\right)\cdot\frac{16}{39}=\boxed{2\frac23}\\\\1)\;3\frac13+5\frac79=\frac{10}3+\frac{52}9=\frac{30}9+\frac{52}9=\frac{82}9;\\\\2)\;3\frac56-\frac7{12}+9\frac78=\frac{23}6-\frac7{12}+\frac{79}8=\frac{92}{24}-\frac{14}{24}+\frac{237}{24}=\frac{315}{24}=\frac{105}8;\\\\3)\;\frac{82}9\cdot2\frac14=\frac{82}9\cdot\frac94=\frac{82}4=\frac{41}2;$

$4)\;1\frac1{15}\cdot\frac{105}8=\frac{16}{15}\cdot\frac{105}8=\frac21\cdot\frac71=14;\\\\5)\;\frac{41}2-14=\frac{41}2-\frac{28}2=\frac{13}2\\\\6)\;\frac{13}2\cdot\frac{16}{39}=\frac83=2\frac23$