Помогите решить номер 587,588

[latex]587.\\1)\;7\frac29\cdot8\frac23-7\frac29\cdot6\frac23=7\frac29\cdot\left(8\frac23-6\frac23\right)=7\frac29\cdot2=14\frac49\\2)\;12\frac{35}{44}\cdot4\frac1{10}-8\frac{35}{44}\cdot4\frac1{10}=4\frac1{10}\cdot\left(12\frac{35}{44}-8\frac{35}{44}\right)=4\frac1{10}\cdot4=16\frac4{10}=16\frac25\\3)\;7\frac13\cdot2\frac15+7\frac13\cdot1\frac45=7\frac13\cdot\left(2\frac15+1\frac45\right)=7\frac13\cdot4=28\frac43=29\frac13[/latex] [latex]4)\;\left(-3\frac19\right)\cdot7\frac47+\left(-3\frac19\right)\cdot\left(-2\frac37\right)=\left(-3\frac19\right)\cdot\left(7\frac47-2\frac37\right)=\\=\left(-3\frac19\right)\cdot5\frac17=-\frac{28}9\cdot\frac{36}7=-4\cdot4=-16\\5)\;2\frac67\cdot4\frac25-2\frac67\cdot4=2\frac67\cdot\left(4\frac25-4\right)=\frac{20}7\cdot\frac25=\frac47\cdot2=\frac87=1\frac17[/latex] [latex]6)\;\left(-2\frac37\right)\cdot(-5)+2\frac37\cdot\left(-2\frac23\right)=2\frac37\cdot\left(5-2\frac23\right)=2\frac37\cdot2\frac13=\frac{17}7\cdot\frac73=\\=\frac{17}3=5\frac23\\\\588.\\1)\;7\frac12\cdot\left(-\frac15\right)+\left(-1\frac23\right)\cdot\left(-\frac9{10}\right)-17\frac{29}{30}=-\frac{15}2\cdot\frac15+\frac53\cdot\frac9{10}-17\frac{29}{30}=\\=-\frac32+\frac32-17\frac{29}{30}=-17\frac{29}{30}[/latex] [latex]2)\;\left(-2\frac{13}{25}\right):\left(-2\frac7{10}\right)-17\frac{25}{47}:\left(-17\frac{25}{47}\right)-4\frac35=\frac{63}{25}:\frac{27}{10}+1-4\frac35=\\=\frac{63}{25}\cdot\frac{10}{27}-3\frac35=\frac75\cdot\frac23-\frac{18}5=\frac{14}{15}-\frac{54}{15}=-\frac{40}{15}=-\frac83=-2\frac23[/latex]