Упр.820 ГДЗ Никольский Потапов 9 класс (Алгебра)

б) ((7 3/7)^2-(6/7)^2)/((17 11/14)^2-(11 3/14)^2).

а)

$$ \frac{\left(7\frac{1}{3}\right)^2-\left(2\frac{2}{3}\right)^2}{\left(5\frac{7}{9}\right)^2-\left(4\frac{2}{9}\right)^2} = \frac{\left(\frac{22}{3}\right)^2-\left(\frac{8}{3}\right)^2}{\left(\frac{52}{9}\right)^2-\left(\frac{38}{9}\right)^2} $$

$$ = \frac{\left(\frac{22}{3}-\frac{8}{3}\right)\left(\frac{22}{3}+\frac{8}{3}\right)}{\left(\frac{52}{9}-\frac{38}{9}\right)\left(\frac{52}{9}+\frac{38}{9}\right)} = \frac{\frac{14}{3}\cdot 10}{\frac{14}{9}\cdot \frac{90}{9}} $$

$$ = \frac{14}{3}\cdot \frac{9}{14}\cdot \frac{30}{3}\cdot \frac{9}{90} = 3 $$

б)

$$ \frac{\left(7\frac{3}{7}\right)^2-\left(\frac{6}{7}\right)^2}{\left(17\frac{11}{14}\right)^2-\left(11\frac{3}{14}\right)^2} = \frac{\left(\frac{52}{7}\right)^2-\left(\frac{6}{7}\right)^2}{\left(\frac{249}{14}\right)^2-\left(\frac{157}{14}\right)^2} $$

$$ = \frac{\left(\frac{52}{7}-\frac{6}{7}\right)\left(\frac{52}{7}+\frac{6}{7}\right)}{\left(\frac{249}{14}-\frac{157}{14}\right)\left(\frac{249}{14}+\frac{157}{14}\right)} = \frac{\frac{46}{7}\cdot \frac{58}{7}}{\frac{92}{14}\cdot \frac{406}{14}} $$

$$ = \frac{46}{7}\cdot \frac{14}{92}\cdot \frac{58}{7}\cdot \frac{14}{406} = \frac{2}{7} $$

Ответ

а) $$3$$; б) $$\frac{2}{7}$$.