Упр.498 ГДЗ Муравин 6 класс (Математика)

Задача

1) 1/7•(-7/9)+(-1/2)^2;
2)-3/4•12/7-(-1/7)^2;
3)-3•(-2/3) :5/12 :(-3/2)^2;
4) 0,5:(-1/3)^2•(-2 2/3)•5/6.

Подробный ответ

1) $$\frac{1}{7}\cdot\left(-\frac{7}{9}\right)+\left(-\frac{1}{2}\right)^2$$

$$ \frac{1}{7}\cdot\left(-\frac{7}{9}\right)=-\frac{1}{9}, \qquad \left(-\frac{1}{2}\right)^2=\frac{1}{4} $$
$$ -\frac{1}{9}+\frac{1}{4}=\frac{-4+9}{36}=\frac{5}{36} $$

2) $$-\frac{3}{4}\cdot\frac{12}{7}-\left(-\frac{1}{7}\right)^2$$

$$ -\frac{3}{4}\cdot\frac{12}{7}=-\frac{9}{7}, \qquad \left(-\frac{1}{7}\right)^2=\frac{1}{49} $$
$$ -\frac{9}{7}-\frac{1}{49}=-\frac{63}{49}-\frac{1}{49}=-\frac{64}{49}=-1\frac{15}{49} $$

3) $$-3\cdot\left(-\frac{2}{3}\right):\frac{5}{12}:\left(-\frac{3}{2}\right)^2$$

$$ -3\cdot\left(-\frac{2}{3}\right)=2, \qquad \left(-\frac{3}{2}\right)^2=\frac{9}{4} $$
$$ 2:\frac{5}{12}:\frac{9}{4} =2\cdot\frac{12}{5}\cdot\frac{4}{9} =\frac{96}{45} =\frac{32}{15} =2\frac{2}{15} $$

4) $$0{,}5:\left(-\frac{1}{3}\right)^2\cdot\left(-2\frac{2}{3}\right)\cdot\frac{5}{6}$$

$$ 0{,}5=\frac{1}{2}, \qquad \left(-\frac{1}{3}\right)^2=\frac{1}{9}, \qquad -2\frac{2}{3}=-\frac{8}{3} $$
$$ \frac{1}{2}:\frac{1}{9}\cdot\left(-\frac{8}{3}\right)\cdot\frac{5}{6} =\frac{1}{2}\cdot 9\cdot\left(-\frac{8}{3}\right)\cdot\frac{5}{6} $$
$$ =\frac{9\cdot 8\cdot 5}{2\cdot 3\cdot 6}\cdot(-1) =-10 $$