№ 9.14 Алгебра = № 19.14 Математика
Обчисліть:
1. $ \ 81\cdot3^{-5};$
2. $ \ -25\cdot{10}^{-2};$
3. $\ 27\cdot\left(-18\right)^{-1};$
4. $\ 2\frac{1}{5}\cdot\left(-\frac{1}{5}\right)^{-1};$
5. $\ -8\cdot2^{-4}+3^0;$
6. $\ 8^{-2}+6^{-1};$
7. $\ {2{,}5}^{-1}+\left(-13\right)^0;$
8. $\ 4^{-3}-\left(-4\right)^{-2};$
9. $\left(-8\right)^{-2}+\left(0,4\right)^{-1};$
10. $\left(\frac{1}{8}\right)^{-2}\cdot{10}^{-3};$
11. $\left(\frac{2}{7}\right)^{-3}:\left(\frac{4}{7}\right)^{-2};$
12. $\ {1{,}25}^{-2}+{2{,}5}^{-3}.$
Розв'язок:
1. $ \ 81\cdot3^{-5}=3^4\cdot3^{-5}=$
$= 3^{-1}=\frac{1}{3};$
2. $ \ -25\cdot{10}^{-2}=-\frac{25}{{10}^2}=$
$= -\frac{25}{100}=-\frac{1}{4};$
3. $\ 27\cdot\left(-18\right)^{-1}=27\cdot\left(-\frac{1}{18}\right)=$
$= -\frac{27}{18}=-\frac{3}{2}=-1{,}5;$
4. $\ 2\frac{1}{5}\cdot\left(-\frac{1}{5}\right)^{-1}=$
$= \frac{11}{5}\cdot\left(-5\right)^1=-\frac{11\cdot5}{5}=-11;$
5. $\ -8\cdot2^{-4}+3^0=-\frac{8}{2^4}+1=$
$= -\frac{8}{16}+1= -\frac{1}{2}+1=$
$= \frac{1}{2}=0{,}5;$
6. $\ 8^{-2}+6^{-1}=\frac{1}{8^2}+\frac{1}{6}=$
$= \frac{1}{64}+\frac{1}{6}=\frac{3}{192}+\frac{32}{192}=$
$= \frac{35}{192};$
7. $\ {2,5}^{-1}+\left(-13\right)^0=$
$= \left(2\frac{5}{10}\right)^{-1}+1=$
$= \left(2\frac{1}{2}\right)^{-1}+1=$
$= \left(\frac{5}{2}\right)^{-1}+1=$
$= \frac{2}{5}+1=0{,}4+1=1{,}4;$
8. $\ 4^{-3}-\left(-4\right)^{-2}=\frac{1}{4^3}-\frac{1}{\left(-4\right)^2}=$
$= \frac{1}{64}-\frac{1}{16}=\frac{1-4}{64}=-\frac{3}{64};$
9. $\left(-8\right)^{-2}+\left(0,4\right)^{-1}=$
$= \frac{1}{\left(-8\right)^2}+\left(\frac{2}{5}\right)^{-1}=$
$= \frac{1}{64}+\frac{5}{2}=\frac{1+160}{64}=\frac{161}{64}=$
$= 2\frac{33}{64};$
10. $\left(\frac{1}{8}\right)^{-2}\cdot{10}^{-3}=\left(\frac{8}{1}\right)^2\cdot\frac{1}{{10}^3}=$
$= \frac{64}{1000}=0{,}064;$
11. $\left(\frac{2}{7}\right)^{-3}:\left(\frac{4}{7}\right)^{-2}=$
$= \left(\frac{7}{2}\right)^3:\left(\frac{7}{4}\right)^2=\frac{7^3}{2^3}:\frac{7^2}{4^2}=$
$= \frac{7^3}{2^3}\cdot\frac{4^2}{7^2}=\frac{7^3\cdot16}{8\cdot7^2}=$
$= \frac{7\cdot2}{1}=14;$
12. $\ {1{,}25}^{-2}+{2{,}5}^{-3}=$
$= \left(\frac{5}{4}\right)^{-2}+\left(\frac{5}{2}\right)^{-3}=$
$= \left(\frac{4}{5}\right)^2+\left(\frac{2}{5}\right)^3=$
$= \frac{80+8}{125}=\frac{88}{125}.$