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Question 1

Question 2

Решите задачу:

$1.\;\frac14\cdot4\frac1{23}\cdot\left(29\frac5{36}-2\frac7{12}-25\frac{14}{15}+1\frac{17}{45}\right):\frac3{23}\\1)\;29\frac5{36}-2\frac7{12}=29\frac5{36}-2\frac{21}{36}=\frac{1049}{36}-\frac{93}{36}=\frac{956}{36}=\frac{239}{9}\\2)\;\frac{239}{9}-25\frac{14}{15}=\frac{239}9-\frac{389}{15}=\frac{1195}{45}-\frac{1167}{45}=\frac{28}{45}\\3)\;\frac{28}{45}+1\frac{17}{45}=\frac{28}{45}+\frac{62}{45}=\frac{90}{45}=2\\4)\;\frac14\cdot4\frac1{23}=\frac14\cdot\frac{93}{23}=\frac{93}{92}$
$5)\;\frac{93}{92}\cdot2=\frac{93}{46}\\6)\;\frac{93}{46}:\frac3{23}=\frac{93}{46}\cdot\frac{23}3=\frac{31}2=15\frac12$

$2.\;\frac2{11}\cdot36\frac14\cdot4\frac25-39:4+68\frac16-70\frac34\\1)\;\frac2{11}\cdot36\frac14=\frac2{11}\cdot\frac{145}4=\frac{145}{22}\\2)\;\frac{145}{22}\cdot4\frac25=\frac{145}{22}\cdot\frac{22}5=29\\3)\;39:4=39\cdot\frac14=\frac{39}4\\4)\;29-\frac{39}4=\frac{116-39}4=\frac{77}4\\5)\;\frac{77}4+68\frac16=\frac{77}4+\frac{409}6=\frac{231}{12}+\frac{818}{12}=\frac{1049}{12}\\6)\;\frac{1049}{12}-70\frac34=\frac{1049}{12}-\frac{283}4=\frac{1049}{12}-\frac{849}4=\frac{200}4=50$

$3.\;90\frac4{17}-3\frac4{51}\cdot3-\frac{53}{99}:\left(82\frac16-81\frac12\right)-68\frac{27}{44}+6\frac16\\1)\;82\frac16-81\frac12=\frac{493}6-\frac{163}2=\frac{493}6-\frac{489}6=\frac46=\frac23\\2)\;3\frac4{51}\cdot3=\frac{157}{51}\cdot7=\frac{157}{17}\\3)\;\frac{53}{99}:\frac23=\frac{53}{99}\cdot\frac32=\frac{53}{66}\\4)\;90\frac4{17}-\frac{157}{17}=\frac{1534}{17}-\frac{157}{17}=\frac{1377}{17}=81\\5)\;81-\frac{53}{66}=80\frac{13}{66}=\frac{5293}{66}$
$6)\;\frac{5293}{66}-68\frac{27}{44}=\frac{5293}{66}-\frac{3019}{44}=\frac{10586}{132}-\frac{9057}{132}=\frac{1529}{132}=\frac{139}{12}\\7)\;\frac{139}{12}+6\frac16=\frac{139}{12}+\frac{37}6=\frac{139}{12}+\frac{74}{12}=\frac{213}{12}=17\frac9{12}=17\frac34$

$4.\;3\frac{11}{12}+\left(84\frac1{48}-80,25\right):0,25-\left(77\frac7{16}-70\right)\cdot\frac2{17}\cdot\frac8{35}\\1)\;84\frac1{48}-80,25=84\frac1{48}-80\frac14=\frac{4033}{48}-\frac{321}4=\frac{4033}{48}-\frac{3852}{48}=\frac{181}{48}\\2)\;77\frac7{16}-70=7\frac7{16}=\frac{119}{16}\\3)\;\frac{181}{48}:0,25=\frac{181}{48}:\frac14=\frac{181}{48}\cdot4=\frac{181}{12}\\4)\;\frac{119}{16}\cdot\frac2{17}=\frac74\\5)\;\frac74\cdot\frac8{35}=\frac25$
$6)\;3\frac{11}{12}+\frac{181}{12}=\frac{47}{12}+\frac{181}{12}=\frac{228}{12}=19\\7)\;19-\frac25=18\frac35$