Упр.853 ГДЗ Никольский Потапов 7 класс (Алгебра)

б) (2*3/4:1,1+3*1/3)/(2,5-0,4*3*1/3) : 5/7 — ((2*1/6+4,5)*0,375)/(2,75 — 1*1/2).

а)

$$ \frac{0{,}5+\frac14+\frac16+0{,}125}{\frac13+0{,}4+\frac{14}{15}} +\frac{(3{,}75-0{,}625)\cdot \frac{48}{125}}{12{,}8\cdot 0{,}25} $$
$$ =\frac{\frac12+\frac14+\frac16+\frac18}{\frac13+\frac25+\frac{14}{15}} +\frac{\frac{25}{8}-\frac58}{\frac{64}{5}\cdot \frac14} $$
$$ =\frac{\frac{25}{24}}{\frac{5}{3}}+\frac{\frac{15}{8}\cdot \frac{48}{125}}{\frac{16}{5}} =\frac{25}{24}\cdot \frac35+\frac{15}{8}\cdot \frac{48}{125}\cdot \frac{5}{16} $$
$$ =\frac{5}{8}+\frac{3}{8}=1. $$

б)

$$ \left(\frac{2\cdot \frac34:1{,}1+3\cdot \frac13}{2{,}5-0{,}4\cdot 3\cdot \frac13}\right):\frac57 -\frac{\left(2\cdot \frac16+4{,}5\right)\cdot 0{,}375}{2{,}75-1\frac12} $$
$$ =\left(\frac{\frac32:\frac{11}{10}+1}{\frac52-\frac25\cdot 1}\right):\frac57 -\frac{\left(\frac13+\frac92\right)\cdot \frac38}{\frac{11}{4}-\frac32} $$
$$ =\left(\frac{\frac32\cdot \frac{10}{11}+1}{\frac52-\frac25}\right):\frac57 -\frac{\frac{29}{6}\cdot \frac38}{\frac54} $$
$$ =\left(\frac{\frac{15}{11}+1}{\frac{21}{10}}\right):\frac57-\frac{29}{16}\cdot \frac45 =\left(\frac{\frac{26}{11}}{\frac{21}{10}}\right):\frac57-\frac{29}{20} $$
$$ =\frac{26}{11}\cdot \frac{10}{21}\cdot \frac75-\frac{29}{20} =\frac{52}{33}-\frac{29}{20} =\frac{1040-957}{660} =\frac{83}{660}. $$

Ответ

а) $$1$$;
б) $$\frac{83}{660}$$.