№ 10.11 Алгебра = № 20.11 Математика
Знайдіть значення виразу:
1. $\ 3^9\cdot3^{-8};$
2. $\ 2^{-3}\cdot2;$
3. $\left(\frac{1}{7}\right)^{-6}\cdot\left(\frac{1}{7}\right)^5;$
4. $\left(\frac{1}{3}\right)^{-9}\cdot\left(\frac{1}{3}\right)^7;$
5. $\ {10}^4\ ∶{10}^5;$
6. $\ 8^{-12}\ ∶8^{-13};$
7. $\ 7\ ∶7^{-1};$
8. $\left(\frac{2}{7}\right)^{-7}:\left(\frac{2}{7}\right)^{-7};$
9. $\left(3^{-1}\right)^4;$
10. $\left(\left(\frac{1}{5}\right)^{-2}\right)^{-1};$
11. $\left({0,2}^3\right)^{-1};$
12. $\left(\left(\frac{7}{13}\right)^0\right)^{-12}.$
Розв'язок:
1. $\ 3^9\cdot3^{-8}=3^{9+\left(-8\right)}=3^1=3;$
2. $\ 2^{-3}\cdot2=2^{-3+1}=2^{-2}=\frac{1}{4};$
3. $\left(\frac{1}{7}\right)^{-6}\cdot\left(\frac{1}{7}\right)^5=$
$= \left(\frac{1}{7}\right)^{-6+5}=\left(\frac{1}{7}\right)^{-1}=7^1=7;$
4. $\left(\frac{1}{3}\right)^{-9}\cdot\left(\frac{1}{3}\right)^7=$
$= \left(\frac{1}{3}\right)^{-9+7}=\left(\frac{1}{3}\right)^{-2}=3^2=9;$
5. $\ {10}^4\ ∶{10}^5={10}^{4-5}={10}^{-1}=$
$= \frac{1}{10}=0,1;$
6. $\ 8^{-12}\ ∶8^{-13}=8^{-12-\left(-13\right)}=$
$= 8^{-12+13}=8^1=8;$
7. $\ 7\ ∶7^{-1}=\frac{7}{1}:\frac{1}{7}=\frac{7}{1}\cdot\frac{7}{1}=$
$= 49;$
8. $\left(\frac{2}{7}\right)^{-7}:\left(\frac{2}{7}\right)^{-7}=$
$= \left(\frac{2}{7}\right)^{-7-\left(-7\right)}=\left(\frac{2}{7}\right)^{-7+7}=$
$= \left(\frac{2}{7}\right)^0=1;$
9. $\left(3^{-1}\right)^4=3^{-4}=\frac{1}{3^4}=\frac{1}{81};$
10. $\left(\left(\frac{1}{5}\right)^{-2}\right)^{-1}=\left(\frac{1}{5}\right)^2=\frac{1}{25};$
11. $\left({0,2}^3\right)^{-1}={0,2}^{-3}=\left(\frac{1}{5}\right)^{-3}=$
$= 5^3=125;$
12. $\left(\left(\frac{7}{13}\right)^0\right)^{-12}=\left(\frac{7}{13}\right)^0=1.$