Упр.534 ГДЗ Муравин 6 класс (Математика)

Задача

1) (55/57-51/95) :(-122/133);
2) (-60/259) :(-45/148+75/222);
3) (8/225-4/105) :(6/35-2/45);
4) (4/51-1/136) :(-5/34-7/51).

Подробный ответ

1) $$\left(\frac{55}{57}-\frac{51}{95}\right):\left(-\frac{122}{133}\right)$$

$$ \frac{55}{57}-\frac{51}{95}=\frac{55\cdot 5-51\cdot 3}{285}=\frac{275-153}{285}=\frac{122}{285} $$
$$ \frac{122}{285}:\left(-\frac{122}{133}\right)=\frac{122}{285}\cdot\left(-\frac{133}{122}\right)=-\frac{133}{285}=-\frac{7}{15} $$

2) $$\left(-\frac{60}{259}\right):\left(-\frac{45}{148}+\frac{75}{222}\right)$$

$$ -\frac{45}{148}+\frac{75}{222}=\frac{-45\cdot 3+75\cdot 2}{444}=\frac{-135+150}{444}=\frac{15}{444}=\frac{5}{148} $$
$$ -\frac{60}{259}:\frac{5}{148}=-\frac{60}{259}\cdot\frac{148}{5} $$
Сократим:
$$ \frac{60}{5}=12,\qquad \frac{148}{259}=\frac{4}{7} $$
Тогда
$$ -\frac{60}{259}\cdot\frac{148}{5}=-12\cdot\frac{4}{7}=-\frac{48}{7}=-6\frac{6}{7} $$

3) $$\left(\frac{8}{225}-\frac{4}{105}\right):\left(\frac{6}{35}-\frac{2}{45}\right)$$

$$ \frac{8}{225}-\frac{4}{105}=\frac{8\cdot 7-4\cdot 15}{1575}=\frac{56-60}{1575}=-\frac{4}{1575} $$
$$ \frac{6}{35}-\frac{2}{45}=\frac{6\cdot 9-2\cdot 7}{315}=\frac{54-14}{315}=\frac{40}{315}=\frac{8}{63} $$
Тогда
$$ -\frac{4}{1575}:\frac{8}{63}=-\frac{4}{1575}\cdot\frac{63}{8} $$
Сократим:
$$ \frac{63}{1575}=\frac{1}{25},\qquad \frac{4}{8}=\frac{1}{2} $$
Получаем
$$ -\frac{1}{25}\cdot\frac{1}{2}=-\frac{1}{50} $$

4) $$\left(\frac{4}{51}-\frac{1}{136}\right):\left(-\frac{5}{34}-\frac{7}{51}\right)$$

$$ \frac{4}{51}-\frac{1}{136}=\frac{4\cdot 8-1\cdot 3}{408}=\frac{32-3}{408}=\frac{29}{408} $$
$$ -\frac{5}{34}-\frac{7}{51}=\frac{-5\cdot 3-7\cdot 2}{102}=\frac{-15-14}{102}=-\frac{29}{102} $$
Тогда
$$ \frac{29}{408}:\left(-\frac{29}{102}\right)=\frac{29}{408}\cdot\left(-\frac{102}{29}\right)=-\frac{102}{408}=-\frac{1}{4} $$