\begin{align}
\end{align}
Cvičení - Mocniny s racionálním mocnitelem
Cvičení 2.28
Cvičení 2.29
Cvičení 2.30
Vypočítej:
a) \(\displaystyle 2^{\Large \frac {3} {2}} \cdot 2^{\Large \frac {4} {3}} \cdot 2^{\Large \frac {7} {6}} = \;\)
\(\displaystyle 2^{\Large \left[\frac {3} {2} + \frac {4} {3} + \frac {7} {6} \right]} = \;\)
\(\displaystyle 2^{\Large \left[\frac {9 \,+ \,8 \,+ \,7} {6} \right]} = \;\)
\(\displaystyle 2^{\Large \frac {24} {6}} = \;\)\(\displaystyle 2^4 = \;\)\(\displaystyle 16\)
b) \(\displaystyle \frac {\left(5^{\Large \frac {3} {5}} \right)^{\Large \frac {3} {2}}} {5^{\Large \frac {4} {10}}} = \;\)
\(\displaystyle \frac {5^{\Large \left[\frac {3} {5} \cdot \frac {3} {2}\right]}} {5^{\Large \frac {4} {10}}} = \;\)
\(\displaystyle \frac {5^{\Large \frac {9} {10}}} {5^{\Large \frac {4} {10}}} = \;\)
\(\displaystyle 5^{\Large \left[\frac {9 \,-\, 4} {10}\right]} = \;\)\(\displaystyle 5^{\Large \frac {5} {10}} = \;\)\(\displaystyle 5^{\Large \frac {1} {2}} = \;\)\(\displaystyle \sqrt{5 \;}\)
c) \(\displaystyle \frac {\left(45^{\Large \frac {2} {5}}\right)^{\Large \frac {17} {4}}}
{5^{\Large \frac {7} {10}}} \cdot 9^{\Large - \,\frac {6} {5}} = \;\)
\(\displaystyle \frac {\left(5^{\Large \frac {2} {5}} \cdot 9^{\Large \frac {2} {5}} \right)^{\Large \frac {17} {4}}}
{5^{\Large \frac {7} {10}}} \cdot 9^{\Large -\,\frac {6} {5}} = \;\)
\(\displaystyle \frac {5^{\Large \left[\frac {2} {5} \cdot \frac {17} {4} \right]} \cdot
9^{\Large \left[\frac {2} {5} \cdot \frac {17} {4} \right]}} {5^{\Large \frac {7} {10}}} \cdot 9^{\Large -\, \frac {6} {5}} = \;\)
\(\displaystyle \frac {5^{\Large \frac {17} {10}} \cdot 9^{\Large \frac {17} {10}}} {5^{\Large \frac {7} {10}}}
\cdot 9^{\Large -\, \frac {6} {5}}= \;\)
\(\displaystyle = 5^{\Large \frac {17 \,-\,7} {10}} \cdot 9^{\Large \frac {17 \,-\,12} {10}}= \;\)\(\displaystyle 5^1 \cdot 9^{\Large \frac {1} {2}}= \;\)\(\displaystyle 5 \cdot 3 = \;\)\(\displaystyle 15\)
d) \(\displaystyle \frac {\left(3^{\Large \frac {30} {7}} \cdot 4^{\Large \frac {18} {7}} \right)^{\Large \frac {7} {6}}}
{36^{\Large \frac {5} {2}}} = \;\)
\(\displaystyle \frac {3^{\Large \left[\frac {30} {7} \cdot \frac {7} {6}\right]} \cdot
4^{\Large \left[\frac {18} {7} \cdot \frac {7} {6}\right]}} {\left(3^2 \cdot 4 \right)^{\Large \frac {5} {2}}} = \;\)
\(\displaystyle \frac {3^5 \cdot 4^3}
{3^{\Large \left[\normalsize 2 \cdot \Large \frac {5} {2}\right]} \cdot 4^{\Large \left[\normalsize 1 \cdot \Large \frac {5} {2} \right]}} = \;\)
\(\displaystyle \frac {3^5 \cdot 4^3} {3^5 \cdot 4^{\Large \frac {5} {2}}} = \;\)
\(\displaystyle 3^{5 \,-\, 5} \cdot 4^{\Large \left[\normalsize 3 \,-\, \Large \frac {5} {2}\right]} = \;\)
\(\displaystyle = 3^0 \cdot 4^{\Large \frac {6 \,-\, 5} {2}} = \;\)
\(\displaystyle 1 \cdot 4^{\Large \frac {1} {2}} = \;\)\(\displaystyle 2\)
Cvičení 2.31
Vyjádři ve tvaru jediné odmocniny:
a) \(\displaystyle \sqrt{2 \;} \cdot \sqrt[\large 3 \,]{2^4 \;} \cdot \sqrt[\large 6 \,]{2^4 \;} = \;\)
\(\displaystyle 2^{\Large \frac {1} {2}} \cdot 2^{\Large \frac {4} {3}} \cdot 2^{\Large \frac {4} {6}} = \;\)
\(\displaystyle 2^{\Large \frac {3 \,+\, 8 \,+\, 4} {6}} = \;\)
\(\displaystyle 2^{\Large \frac {15} {6}} = \;\)\(\displaystyle 2^{\Large \frac {5} {2}} = \;\)\(\displaystyle \sqrt{2^5} = \;\)\(\displaystyle \sqrt{32}\)
b) \(\displaystyle \frac {\sqrt[\large 3 \,]{6 \;}} {\sqrt[\large 6 \,]{36 \;}} \cdot \sqrt[\large 4 \,]{6^3 \;} = \;\)
\(\displaystyle \frac {6^{\Large \frac {1} {3}}} {\left(6^2 \right)^{\Large \frac {1} {6}}} \cdot
6^{\Large \frac {3}{4}} = \;\)
\(\displaystyle \frac {\; 6^{\Large \frac {1} {3}}\;} {6^{\Large \frac {2} {6}}} \cdot 6^{\Large \frac {3}{4}} = \;\)
\(\displaystyle 6^{\Large \frac {4 \,+\, 9 \,-\, 4} {12}} = \;\)\(\displaystyle 6^{\Large \frac {9} {12}} = \;\)\(\displaystyle 6^{\Large \frac {3} {4}} = \;\)\(\displaystyle \sqrt[\large 4 \,]{6^3}\)
c) \(\displaystyle \frac {1} {10\,000} \cdot \sqrt[\large 4 \,]{5 \cdot \sqrt[\large 3 \,]{2^4 \cdot 5 \;}\;} = \;\)
\(\displaystyle 10^{-\,4} \cdot \sqrt[\large 4 \,]{5 \cdot 2^{\Large \frac {4} {3}} \cdot 5^{\Large \frac {1} {3}\;}\;} = \;\)
\(\displaystyle 10^{-\,4} \cdot \left(5^{\Large \frac {3 \,+\, 1} {3}} \cdot 2^{\Large \frac {4} {3}} \right)^{\Large \frac {1} {4}} = \;\)
\(\displaystyle 10^{-\,4} \cdot \left(5^{\Large \frac {4} {3}} \cdot 2^{\Large \frac {4} {3}}\right)^{\Large \frac {1} {4}} = \;\)
\(\displaystyle = 10^{-\,4} \cdot 10^{\Large \left[\frac {4} {3} \cdot \frac {1} {4}\right]} = \;\)
\(\displaystyle 10^{-\,4} \cdot 10^{\Large \frac {1} {3}} = \;\)
\(\displaystyle 10^{\Large \frac {-\,12 + 1\;} {3}}= \;\)
\(\displaystyle 10^{\Large \frac {-\,11\;} {3}}= \;\)
\(\displaystyle \sqrt[\large 3 \,]{\frac {1} {10^{11}}}\)
d) \(\displaystyle \frac {\sqrt[\large 5 \,]{144 \;} \cdot \sqrt[\large 10 \,]{4^6 \;}} {\sqrt{27 \;}} \cdot
\sqrt[\large 5 \,]{\sqrt{4 \;}\;} = \;\)
\(\displaystyle \frac {\sqrt[\large 5 \,]{12^2\;} \cdot 4^{\Large \frac {6} {10}}} {\sqrt{3^3 \;}} \cdot
\left(4^{\Large \frac {1} {2}}\right)^{\Large \frac {1} {5}} = \;\)
\(\displaystyle \frac {12^{\Large \frac {2} {5}} \cdot 4^{\Large \frac {3}{5}}} {3^{\Large \frac {3} {2}}}
\cdot 4^{\Large \frac {1} {10}} = \;\)
\(\displaystyle \frac {3^{\Large \frac {2} {5}} \cdot 4^{\Large \frac {2} {5}} \cdot 4^{\Large \frac {3} {5}}}
{3^{\Large \frac {3} {2}}} \cdot 4^{\Large \frac {1} {10}} = \;\)
\(\displaystyle = 3^{\Large \frac {4 \,-\, 15} {10}} \cdot 4^{\Large \frac {4 \,+\, 6 \,+\, 1} {10}}= \;\)
\(\displaystyle 3^{\Large \frac {-\,11\;} {10}} \cdot 4^{\Large \frac {11} {10}} = \;\)
\(\displaystyle \left(\frac {4} {3} \right)^{\Large \frac {11} {10}} = \;\)\(\displaystyle \sqrt[\large 10 \,]{\left(\frac {4} {3} \right)^{11}\;}\)
Cvičení 2.32
Vypočítej:
a) \(\displaystyle \left[5 \cdot \left(5 \cdot 5^4 \right)^{3}\right]^{\Large \frac {3} {4}} = \;\)
\(\displaystyle \left(5 \cdot 5^3 \cdot 5^{[4 \, \cdot 3]} \right)^{\Large \frac {3} {4}} = \;\)
\(\displaystyle \left(5 \cdot 5^3 \cdot 5^{12} \right)^{\Large \frac {3} {4}} = \;\)
\(\displaystyle \left(5^{[1 \,+ \,3 \,+\, 12]} \right)^{\Large \frac {3} {4}} = \;\)
\(\displaystyle = \left(5^{16}\right)^{\Large \frac {3} {4}} = \;\)
\(\displaystyle 5^{\Large \left[\normalsize 16 \cdot \Large \frac {3} {4} \right]}= \;\)\(\displaystyle 5^{12}\)
b) \(\displaystyle 2^{\Large \frac {4} {3}} \cdot \left(\frac {1} {2} \right)^3 \cdot
\left(2^{\Large \frac {3} {2}} \right)^4 = \;\)
\(\displaystyle 2^{\Large \frac {4} {3}} \cdot \left(2^{-\,1} \right)^3 \cdot
2^{\Large \left[\frac {3} {2} \cdot \normalsize 4\right]} = \;\)
\(\displaystyle 2^{\Large \frac {4} {3}} \cdot 2^{-\,3} \cdot 2^6 = \;\)
\(\displaystyle 2^{\Large \frac {4 \,-\, 9 \,+\, 18} {3}} = \;\)\(\displaystyle 2^{\Large \frac {13} {3}}\)
c) \(\displaystyle \left(\frac {4} {27} \right)^{\Large \frac {1} {3}} \div
\left(\frac {32} {81} \right)^{\Large \frac {1} {4}} = \;\)
\(\displaystyle \left(\frac {2^2} {3^3} \right)^{\Large \frac {1} {3}} \div
\left(\frac {2^5} {3^4} \right)^{\Large \frac {1} {4}} = \;\)
\(\displaystyle \frac {2^{\Large \frac {2} {3}}} {3^{1}} \cdot \frac {3^1} {2^{\Large \frac {5} {4}}} = \;\)
\(\displaystyle 2^{\Large \frac {8\,-\,15} {12}} = \;\)\(\displaystyle 2^{\Large -\,\frac {7} {12}}\)
d) \(\displaystyle \frac {3^{\Large \frac {5} {4}} \cdot 8^{\Large \frac {12}{9}} \cdot (0,5)^{\Large \frac {2} {3}}}
{\left(\large \frac {5} {10} \right)^{\Large \frac {6} {18}} \cdot 3^{\Large \frac {1} {4}}} = \;\)
\(\displaystyle \frac {3^{\Large \frac {5} {4}} \cdot \left(2^3\right)^{\Large \frac {4} {3}} \cdot
\left(\large \frac {1} {2} \right)^{\Large \frac {2} {3}}} {\left(\large \frac {1} {2} \right)^{\Large \frac {1} {3}} \cdot
3^{\Large \frac {1} {4}}} = \;\)
\(\displaystyle \frac {3^{\Large \frac {5} {4}} \cdot 2^4 \cdot 2^{\Large -\,\frac {2} {3}}}
{2^{\Large -\,\frac {1} {3}} \cdot 3^{\Large \frac {1} {4}}} = \;\)
\(\displaystyle 3^{\Large \frac {5 \,-\,1} {4}} \cdot 2^{\Large \frac {12 \,-\,2\,-\,(\,-1)} {3}} = \;\)\(\displaystyle 3 \cdot 2^{\Large \frac {11} {3}}\)
Cvičení 2.33 
Zjednoduš za předpokladu, že \(a > 0\), \(b > 0\):
a) \(\displaystyle \frac {a^{0,2} \cdot \left(\large \frac {1} {a} \right)^{\Large \frac {14} {8}} \cdot
a^{1,5}} {a^{0,5} \cdot \left(a^{\Large \frac {1} {3}} \right)^4} = \;\)
\(\displaystyle \frac {a^{\Large \frac {1} {5}} \cdot a^{\Large -\,\frac {7} {4}} \cdot a^{\Large \frac {3} {2}}}
{a^{\Large \frac {1} {2}} \cdot a^{\Large \frac {4} {3}}} = \;\)
\(\displaystyle \frac {a^{\Large \frac {4 \,- \,35 \,+\, 30} {20}}} {a^{\Large \frac {3 \,+\,8} {6}}} = \;\)
\(\displaystyle \frac {a^{\Large -\,\frac {1} {20}}} {a^{\Large \frac {11} {6}}} = \;\)
\(\displaystyle a^{\Large \frac {-\, 3 \,-\, 110} {60}} = \;\)
\(\displaystyle a^{\Large \frac {-113} {60}}\)
b) \(\displaystyle \left[\frac {a^{\Large \frac {4} {3}} \cdot b^{0,75}}
{\left(a^2\right)^{0,8}}\right]^{2,5} = \;\)
\(\displaystyle \left(\frac {a^{\Large \frac {4} {3}} \cdot b^{\Large \frac {3} {4}}}
{a^{\Large \frac {8}{5}}} \right)^{\Large \frac {5} {2}} = \;\)
\(\displaystyle \left(a^{\Large \frac {20 \,-\, 24} {15}} \cdot b^{\Large \frac {3} {4}} \right)^{\Large \frac {5} {2}} = \;\)
\(\displaystyle \left(a^{\Large -\, \frac {4} {15}} \cdot b^{\Large \frac {3} {4}} \right)^{\Large \frac {5} {2}} = \;\)
\(\displaystyle a^{\Large -\,\frac {2} {3}} \cdot b^{\Large \frac {15} {8}}\)
Cvičení 2.34
Za předpokladu, že \(a > 0\), vyjádři ve tvaru jediné odmocniny:
a) \(\displaystyle \frac{1}{a} \cdot \sqrt[\large 7 \,]{a^6 \cdot \sqrt[\large 5 \,]{a^4 \cdot \sqrt{a^6 \;}\;}\;} = \;\)
\(\displaystyle a^{-\,1} \cdot \left[a^6 \cdot \left(a^4 \cdot a^{\Large \frac {6} {2}} \right)^{\Large \frac {1} {5}} \right]^{\Large \frac {1} {7}} = \;\)
\(\displaystyle a^{-\,1} \cdot a^{\Large \frac {6} {7}} \cdot \left(a^{\Large \frac {8\,+\,6}{2}} \right)^{\Large \frac {1} {35}} = \;\)
\(\displaystyle = a^{\Large \frac {-\,7 \,+\, 6} {7}} \cdot \left(a^{7}\right)^{\Large \frac {1} {35}} = \;\)
\(\displaystyle a^{\Large -\,\frac {1} {7}} \cdot a^{\Large \frac {7} {35}} = \;\)
\(\displaystyle a^{\Large \frac {-\,5 \,+\,7} {35}} = \;\)
\(\displaystyle a^{\Large \frac {2} {35}} = \;\)\(\displaystyle \sqrt[\large 35 \,]{a^{2}\;}\)
b) \(\displaystyle \frac {\sqrt[\large 4 \,]{a^5 \cdot \sqrt[\large 3 \,]{a^2 \;}\;}} {\sqrt{a^9 \cdot \sqrt[\large 3 \,]{a \;}\;}} = \;\)
\(\displaystyle \frac {\left(a^5 \cdot a^{\Large \frac {2} {3}} \right)^{\Large \frac {1} {4}}}
{\left(a^9 \cdot a^{\Large \frac {1} {3}} \right)^{\Large \frac {1} {2}}} = \;\)
\(\displaystyle \frac {a^{\Large \frac {5} {4}} \cdot a^{\Large \frac {1} {6}}}
{a^{\Large \frac {9} {2}} \cdot a^{\Large \frac {1} {6}}} = \;\)
\(\displaystyle a^{\Large \frac {5 \,-\, 18} {4}} = \;\)
\(\displaystyle a^{\Large -\,\frac {13} {4}} = \;\)
\(\displaystyle \sqrt[\large 4 \,]{\frac {1} {a^{13}}\;}\)
