Упр.22.9 ГДЗ Мордкович 7 класс (Алгебра)

а) (5/2)2 :(-25/4)*(5/2)0;
б) (1/3)*(-1/9):(1/3)5;
в) 1,5^4 : (-1,5)3 * (-1,5)0 : 1,5;
г) (8/27): (2/3)2*(16/81)0.

1. $$\left(\frac{5}{2}\right)^2 : \left(-\frac{25}{4}\right)\cdot \left(\frac{5}{2}\right)^0$$

   $$\left(\frac{5}{2}\right)^2=\frac{25}{4}, \qquad \left(\frac{5}{2}\right)^0=1$$

   $$\frac{25}{4} : \left(-\frac{25}{4}\right)\cdot 1=-1$$

2. $$\left(\frac{1}{3}\right)^3\cdot \left(-\frac{1}{9}\right):\left(\frac{1}{3}\right)^5$$

   $$\left(\frac{1}{3}\right)^3=\frac{1}{3^3}, \qquad -\frac{1}{9}=-\frac{1}{3^2}$$

   $$\frac{1}{3^3}\cdot \left(-\frac{1}{3^2}\right):\frac{1}{3^5} =\frac{1}{3^3}\cdot \left(-\frac{1}{3^2}\right)\cdot 3^5=-1$$

3. $$1{,}5^4 : (-1{,}5)^3 \cdot (-1{,}5)^0 : 1{,}5$$

   $$(-1{,}5)^3=-1{,}5^3,\qquad (-1{,}5)^0=1$$

   $$1{,}5^4 : (-1{,}5)^3 \cdot 1 : 1{,}5=-1{,}5\cdot 1 : 1{,}5=-1$$

4. $$\frac{8}{27} : \left(\frac{2}{3}\right)^2 \cdot \left(\frac{16}{81}\right)^0$$

   $$\left(\frac{2}{3}\right)^2=\frac{4}{9}, \qquad \left(\frac{16}{81}\right)^0=1$$

   $$\frac{8}{27} : \frac{4}{9}\cdot 1=\frac{8}{27}\cdot \frac{9}{4}=\frac{2}{3}$$

Ответ

а) $$-1$$; б) $$-1$$; в) $$-1$$; г) $$\frac{2}{3}$$.