Упр.92 ГДЗ Алимов 10-11 класс (Алгебра)

Вычислить:

1) $$\left(0{,}645 : 0{,}3 — 1 * \frac{107}{180}\right) * \left(4 : 6{,}25 — 1 : 5 + \frac{1}{7} * 1{,}96$$;

2) $$\left(\frac{1}{2} — 0{,}374\right) : 0{,}125 + \left(\frac{5}{6} — \frac{7}{12}\right) : \left(0{,}358 — 0{,}108\right)$$.

1. $$\left(0{,}645:0{,}3-1\frac{107}{180}\right)\cdot\left(4:6{,}25-1:5+\frac17\cdot1{,}96\right)=$$

   $$=\left(\frac{645}{1000}:\frac{3}{10}-\frac{180+107}{180}\right)\cdot\left(\frac{4}{1}:\frac{625}{100}-\frac15+\frac17\cdot\frac{196}{100}\right)=$$

   $$=\left(\frac{129}{200}\cdot\frac{10}{3}-\frac{287}{180}\right)\cdot\left(\frac{4\cdot100}{625}-\frac15+\frac{28}{100}\right)=$$

   $$=\left(\frac{43}{20}-\frac{287}{180}\right)\cdot\left(\frac{16}{25}-\frac15+\frac{7}{25}\right)=$$

   $$=\left(\frac{387}{180}-\frac{287}{180}\right)\cdot\left(\frac{23}{25}-\frac{5}{25}\right)=$$

   $$=\frac{100}{180}\cdot\frac{18}{25}=\frac{5}{9}\cdot\frac{18}{25}=\frac{2}{5}=0{,}4.$$

2. $$\left(\frac12-0{,}375\right):0{,}125+\left(\frac56-\frac{7}{12}\right):(0{,}358-0{,}108)=$$

   $$=\left(\frac12-\frac{375}{1000}\right):\frac{125}{1000}+\left(\frac{10}{12}-\frac{7}{12}\right):\left(\frac{358}{1000}-\frac{108}{1000}\right)=$$

   $$=\left(\frac12-\frac38\right):\frac18+\frac{3}{12}:\frac14=$$

   $$=\frac18:\frac18+\frac14:\frac14=1+1=2.$$

Ответ: 1) $$0{,}4$$; 2) $$2$$.