a) Obliczmy wartość wyrażenia:

 

${\left[3-{\left(3-{\left(3-3^{-1}\right)}^{-1}\right)}^{-1}\right]}^{-1}={\left[3-{\left(3-{\left(3-\frac{1}{3}\right)}^{-1}\right)}^{-1}\right]}^{-1}={\left[3-{\left(3-{\left(\frac{8}{3}\right)}^{-1}\right)}^{-1}\right]}^{-1}=$  

$={\left[3-{\left(3-\frac{3}{8}\right)}^{-1}\right]}^{-1}=={\left[3-{\left(\frac{24}{8}-\frac{3}{8}\right)}^{-1}\right]}^{-1}={\left[3-{\left(\frac{21}{8}\right)}^{-1}\right]}^{-1}={\left[3-\frac{8}{21}\right]}^{-1}=$

$={\left[\frac{63}{21}-\frac{8}{21}\right]}^{-1}={\left(\frac{55}{21}\right)}^{-1}=\frac{21}{55}=0,3\left(81\right)$   

 

Odp. 381

 

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b) Obliczmy wartość poniższego wyrażenia:

$\frac{{\left(1,4\right)}^{-3}\cdot {\left(-\frac{5}{7}\right)}^{-1}}{{\left[{\left(\frac{3}{7}\right)}^3\right]}^{-1}\cdot {\left(-\frac{5}{3}\right)}^{-3}:{\left(\frac{5}{7}\right)}^{-2}}=\frac{{\left(1\frac{2}{5}\right)}^{-3}\cdot \left(-\frac{7}{5}\right)}{{\left(\frac{3^3}{7^3}\right)}^{-1}\cdot \left(-\frac{3^3}{5^3}\right)\cdot \frac{7^2}{5^2}}=$  

  $=\frac{{\left(\frac{7}{5}\right)}^{-3}\cdot \left(-\frac{7}{5}\right)}{\frac{7^3}{{\cancel{3^3}}_1}\cdot \left(-\frac{{\cancel{3^3}}^1}{5^3}\right)\cdot \frac{7^2}{5^2}}=\frac{\frac{{\cancel{5^3}}^{5^2}}{7^3}\cdot \left(-\frac{7}{{\cancel{5}}_1}\right)}{-\frac{7^3}{5^3}\cdot \frac{7^2}{5^2}}=$  

$=\frac{\frac{5^2}{{\cancel{7^3}}^{7^2}}\cdot \left(-{\cancel{7}}^1\right)}{-\frac{7^5}{5^5}}=\frac{-{\left(\frac{5}{7}\right)}^2}{-{\left(\frac{7}{5}\right)}^5}={\left(\frac{5}{7}\right)}^2\cdot {\left(\frac{5}{7}\right)}^5={\left(\frac{5}{7}\right)}^7=\frac{78125}{823543}\approx 0,0949$  

 

Odp. 094

 

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c) Obliczmy wartość wyrażenia:

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

$=\frac{{10}^4\cdot \left(26+3\frac{1}{2}\right)}{2\frac{4}{5}\cdot {10}^3:\left(\frac{16}{5}\cdot \frac{1}{{10}^3}\right)}=\frac{{10}^4\cdot 29\frac{1}{2}}{\frac{14}{{\cancel{5}}_1}\cdot {10}^3\cdot \frac{{\cancel{5}}^1}{16}\cdot {10}^3}=$  

$=\frac{{10}^4\cdot \frac{59}{2}}{{10}^6\cdot \frac{{\cancel{14}}^7}{{\cancel{16}}_8}}=\frac{1}{{\cancel{100}}_{25}}\cdot \frac{59}{2}\cdot \frac{{\cancel{8}}^2}{7}=\frac{59\cdot {\cancel{2}}^1}{25\cdot {\cancel{2}}_1\cdot 7}=\frac{59}{175}=0,33\left(714285\right)$  

 

Odp. 337