Упр.545 ГДЗ Никольский Потапов 6 класс (Математика)

а) 1/2•(-2/3)+(-1/2)^2;
б) -3/4•12/7-(-1/7)^2;
в) -1/3•6/5-5/6•3/25;
г) 3/10•(-5/6)+2/3•(-3/8).

а) $$\frac{1}{2}\cdot\left(-\frac{2}{3}\right)+\left(-\frac{1}{2}\right)^2$$

$$ \frac{1}{2}\cdot\left(-\frac{2}{3}\right)=-\frac{1}{3}, \qquad \left(-\frac{1}{2}\right)^2=\frac{1}{4} $$
$$ -\frac{1}{3}+\frac{1}{4}=\frac{-4+3}{12}=-\frac{1}{12} $$

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

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

в) $$-\frac{1}{3}\cdot\frac{6}{5}-\frac{5}{6}\cdot\frac{3}{25}$$

$$ -\frac{1}{3}\cdot\frac{6}{5}=-\frac{2}{5}, \qquad \frac{5}{6}\cdot\frac{3}{25}=\frac{1}{10} $$
$$ -\frac{2}{5}-\frac{1}{10}=-\frac{4}{10}-\frac{1}{10}=-\frac{5}{10}=-\frac{1}{2} $$

г) $$\frac{3}{10}\cdot\left(-\frac{5}{6}\right)+\frac{2}{3}\cdot\left(-\frac{3}{8}\right)$$

$$ \frac{3}{10}\cdot\left(-\frac{5}{6}\right)=-\frac{1}{4}, \qquad \frac{2}{3}\cdot\left(-\frac{3}{8}\right)=-\frac{1}{4} $$
$$ -\frac{1}{4}-\frac{1}{4}=-\frac{1}{2} $$

Ответ

а) $$-\frac{1}{12}$$; б) $$-1\frac{15}{49}$$; в) $$-\frac{1}{2}$$; г) $$-\frac{1}{2}$$.