\begin{align}
\end{align}
Cvičení - Odmocniny z reálného čísla
Cvičení 2.19
Cvičení 2.20
Vypočítej:
a) \(\displaystyle \sqrt[\large 6 \,]{27} \cdot \sqrt[\large 6]{9 \;} =\)
\(\displaystyle \sqrt[\large 6 \,]{3^3 \;} \cdot \sqrt[\large 6 \,]{3^2 \;} = \;\)\(\displaystyle \sqrt[\large 6 \,]{3^3 \cdot 3^2 \;} = \;\)\(\displaystyle \sqrt[\large 6 \,]{3^5 \;}\)\(\displaystyle 3\)
b) \(\displaystyle \frac {\sqrt[\large 3 \;]{24 \;}} {\sqrt[\large 3 \,]{3} \;} =\)
\(\displaystyle \sqrt[\large 3 \,]{\frac {24} {3} \;} = \;\)\(\displaystyle \sqrt[\large 3 \,]{8 \;} = \;\)\(\displaystyle 2\)
c) \(\displaystyle \sqrt[\large 3 \,]{25 \;} \cdot \sqrt[\large 3 \,]{- \,5 \;} =\)
Odmocnina ze záporného čísla není definována.
d) \(\displaystyle \sqrt[\large 4 \,]{\frac {81} {16} \;} \cdot \sqrt[\large 3 \,]{0,008 \;} =\)
\(\displaystyle \frac {\sqrt[\large 4 \,]{81 \;}} {\sqrt[\large 4 \,]{16} \;} \cdot \sqrt[\large 3 \,]{0,008 \;} = \;\)\(\displaystyle \frac {3} {2} \cdot 0,2 = \;\)\(\displaystyle 0,3\)
Cvičení 2.21
Rozhodni, zda pro \(a > 0\) platí:
Cvičení 2.22
Vypočítej:
a) \(\displaystyle \left(\sqrt[\large 3 \,]{3 \;} \right)^4 =\)
\(\displaystyle \left(\sqrt[\large 3 \,]{3^4 \;}\right) = \;\)\(\displaystyle \left(\sqrt[\large 3 \,]{3^3 \cdot 3^1 \;} \right) = \;\)\(\displaystyle \sqrt[\large 3 \,]{3^3 \;} \cdot \sqrt[\large 3 \,]{3 \;} = \;\)\(\displaystyle 3 \sqrt[\large 3 \,]{3 \;} \)
b) \(\displaystyle \left(\sqrt[\large 3 \,]{\frac {2} {3^3} \;} \right)^5 =\)
\(\displaystyle \sqrt[\large 3 \,]{\left(\frac {\, 2 \,}{3^3}\right)^5 \;} = \;\)
\(\displaystyle \sqrt[\large 3 \,]{\frac {\, 2^5 \,} {\left(3^3\right)^5} \;} = \;\)
\(\displaystyle \frac {\sqrt[\large 3 \,]{2^3 \cdot 2^2} \;} {\sqrt[\large 3 \,]{\left(3^5 \right)^3} \;} = \;\)
\(\displaystyle \frac {\sqrt[\large 3 \,]{2^3 \;} \cdot \sqrt[\large 3 \,]{4}} {3^5 \;} = \;\)\(\displaystyle \frac {2 \cdot \sqrt[\large 3 \,]{4 \;}} {3^5}\)
c) \(\displaystyle \sqrt[\large 3 \,]{\sqrt[\large 3 \,]{1 \;} \;} =\)
\(\displaystyle \sqrt[\large 3 \cdot 3 \,]{1 \;} = \;\)\(\displaystyle \sqrt[\large 9 \,]{1 \;} = \;\)\(\displaystyle 1\)
d) \(\displaystyle \sqrt[\large 6 \,]{81 \;} \cdot \sqrt{\sqrt[\large 3 \,]{3^2 \;} \;} =\)
\(\displaystyle \sqrt[\large 6 \,]{3^4 \;} \cdot \sqrt[\large 2 \cdot 3 \,]{3^2 \;} = \;\)
\(\displaystyle \sqrt[\large 6 \,]{3^4 \cdot 3^2 \;} = \;\)\(\displaystyle \sqrt[\large 6 \,]{3^6 \;} = \;\)\(\displaystyle 3\)
e) \(\displaystyle \frac {\sqrt[\large 12 \,]{2^6 \;}} {\sqrt[\large 20 \,]{16^5 \;}} =\)
\(\displaystyle \frac {\sqrt[\large 6 \cdot 2 \,]{2^6}} {\sqrt[\large 5 \cdot \, 4 \,]{16^5 \;}} = \;\)
\(\displaystyle \frac {\sqrt{2}} {\sqrt[\large 4 \,]{16 \;}} = \;\)\(\displaystyle \frac {\sqrt{2}} {\sqrt[\large 4 \,]{2^4 \;}} = \;\)\(\displaystyle \frac {\sqrt{2 \;}} {2}\)
Cvičení 2.23
Usměrni zlomky:
a) \(\displaystyle \frac {6} {\sqrt{6 \;}} =\)
\(\displaystyle \frac {6} {\sqrt{6 \;}} \cdot \frac {\sqrt{6 \;}} {\sqrt{6 \;}} = \;\)\(\displaystyle \frac {6 \cdot \sqrt{6 \;}} {6} = \;\)\(\displaystyle \sqrt{6 \;}\)
b) \(\displaystyle \frac {5} {\sqrt[\large 3 \,]{3 \;}} =\)
\(\displaystyle \frac {5} {\sqrt[\large 3 \,]{3 \;}} \cdot
\frac {\sqrt[\large 3 \,]{3 \;} \cdot \sqrt[\large 3 \,]{3}} {\sqrt[\large 3 \,]{3 \;} \cdot \sqrt[\large 3 \,]{3}} = \;\)
\(\displaystyle \frac {5 \cdot \sqrt[\large 3 \,]{9 \;}} {3}\)
c) \(\displaystyle \frac {\sqrt{5 \;} + \sqrt{2 \;}} {\sqrt{5 \;} - \sqrt{2 \;}} =\)
\(\displaystyle \frac {\sqrt{5 \;} + \sqrt{2 \;}} {\sqrt{5 \;} - \sqrt{2 \;}} \cdot
\frac {\sqrt{5} + \sqrt{2}} {\sqrt{5} + \sqrt{2}} = \;\)
\(\displaystyle \frac {\sqrt{5 \;}\sqrt{5 \;} + \sqrt{2 \;}\sqrt{5 \;} + \sqrt{2 \;}\sqrt{5 \;} + \sqrt{2 \;}\sqrt{2 \;}}
{\left(\sqrt{5 \;} \right)^2 - \left(\sqrt{2 \;} \right)^2} = \;\)
\(= \displaystyle \frac {5 + 2\sqrt{2 \;}\sqrt{5 \;} + 2} {5 - 2} = \;\)\(\displaystyle \frac {7 + 2\sqrt{2 \;}\sqrt{5 \;}} {3} = \;\)\(\displaystyle \frac {7 + 2\sqrt{10 \;}} {3}\)
d) \(\displaystyle \frac {\sqrt{2 \;} \cdot \sqrt{5 \;}} {\sqrt{10 \;} + 3} =\)
\(\displaystyle \frac {\sqrt{2 \;} \cdot \sqrt{5 \;}} {\sqrt{10 \;} + 3} \cdot \frac {\sqrt{10 \;} - 3} {\sqrt{10 \;} - 3} = \;\)
\(\displaystyle \frac {\sqrt{2}\sqrt{5}\sqrt{10} - 3 \cdot \sqrt{2}\sqrt{5}} {\left(\sqrt{10} \right)^2 - 3^2} = \;\)
\(\displaystyle \frac {\sqrt{10 \;} \sqrt{10 \;} - 3 \cdot \sqrt{10 \;}} {10 - 9} = \;\)
\(\displaystyle = \frac {10 - 3 \cdot \sqrt{10 \;}} {1} = \;\)\(\displaystyle 10 - 3 \cdot \sqrt{10 \;}\)
Cvičení 2.24
Vyjádři ve tvaru jediné odmocniny:
a) \(\displaystyle \sqrt[\large 3 \,]{2 \;} \cdot \sqrt{3 \;} =\)
\(\displaystyle \sqrt[\large 3 \cdot 2 \,]{2^2 \;} \cdot \sqrt[\large 2 \cdot 3 \,]{3^3 \;} = \;\)
\(\displaystyle \sqrt[\large 6 \,]{2^2 \cdot 3^3 \;} = \;\)\(\displaystyle \sqrt[\large 6 \,]{4 \cdot 27 \;} = \;\)\(\displaystyle \sqrt[\large 6]{108 \;}\)
b) \(\displaystyle 3\sqrt[\large 3 \,]{3 \;} =\)
\(\displaystyle \sqrt[\large 3 \,]{3^3 \;} \cdot \sqrt[\large 3 \,]{3 \;} = \;\)\(\displaystyle \sqrt[\large 3 \,]{3^3 \cdot 3} = \;\)\(\displaystyle \sqrt[\large 3 \,]{81}\)
c) \(\displaystyle \frac {\sqrt[\large 4 \,]{4 \;}} {\sqrt[\large 3 \,]{2 \;}} =\)
\(\displaystyle \frac {\sqrt[\large 4 \cdot 3 \,]{4^3 \;}} {\sqrt[\large 3 \cdot \, 4 \,]{2^4 \;}} = \;\)\(\displaystyle \sqrt[\large 12 \,]{\frac{64}{16} \;} = \;\)\(\displaystyle \sqrt[\large 12 \,]{4 \;}\)
d) \(\displaystyle \frac {\sqrt{10 \;}} {\sqrt[\large 3 \,]{100 \;}} =\)
\(\displaystyle \frac {\sqrt[\large 2 \cdot 3 \,]{10^3 \;}} {\sqrt[\large 3 \cdot 2 \,]{100^2 \,} \;} = \;\)
\(\displaystyle \sqrt[\large 6 \,]{\frac {1\,000} {10\,000} \;} = \;\)\(\displaystyle \sqrt[\large 6 \,]{\frac {1} {10} \;}\)
Cvičení 2.25
Vypočítej:
a) \(\displaystyle \sqrt[\large 3 \,]{16 \;} + \sqrt[\large 3 \,]{250 \;} =\)
\(\displaystyle \sqrt[\large 3 \,]{2^4 \;} + \sqrt[\large 3 \,]{5^3 \cdot 2 \;} = \;\)
\(\displaystyle \sqrt[\large 3 \,]{2^3} \cdot \sqrt[\large 3 \,]{2} + \sqrt[\large 3 \,]{5^3 \;} \cdot \sqrt[\large 3 \,]{2 \;} = \;\)
\(\displaystyle 2 \cdot \sqrt[\large 3 \,]{2 \;} + 5 \cdot \sqrt[\large 3 \,]{2 \;} = \;\)
\(\displaystyle = (2 + 5) \cdot \sqrt[\large 3 \,]{2 \;} = \;\)\(\displaystyle 7 \cdot \sqrt[\large 3 \,]{2 \;}\)
b) \(\displaystyle \frac {\sqrt[\large 4 \,]{48 \;}} {\sqrt[\large 4 \,]{243 \;}} =\)
\(\displaystyle \frac {\sqrt[\large 4 \,]{2^4 \cdot 3 \;}} {\sqrt[\large 4 \,]{3^5}} = \;\)
\(\displaystyle \frac {\sqrt[\large 4 \,]{2^4 \;} \cdot \sqrt[\large 4 \,]{3 \;}} {\sqrt[\large 4 \,]{3^4 \;} \cdot \sqrt[\large 4 \,]{3 \;}} = \;\)
\(\displaystyle \frac {2 \cdot \sqrt[\large 4 \,]{3 \;}} {3 \cdot \sqrt[\large 4 \,]{3} } = \;\)\(\displaystyle \frac {2} {3}\)
c) \(\displaystyle 6 \cdot \sqrt[\large 3 \,]{270 \;} - 9 \cdot \sqrt[\large 3 \,]{80 \;} =\)
\(\displaystyle 6 \cdot \sqrt[\large 3 \,]{2 \cdot 3^3 \cdot 5 \;} - 9 \cdot \sqrt[\large 3 \,]{2^4 \cdot 5} = \;\)
\(\displaystyle 6 \cdot \sqrt[\large 3 \,]{3^3 \;} \cdot \sqrt[\large 3 \,]{10 \;} -
9 \cdot \sqrt[\large 3 \,]{2^3 \;} \cdot \sqrt[\large 3 \,]{10 \;} = \;\)
\(\displaystyle = 6 \cdot 3 \cdot \sqrt[\large 3 \,]{10} - 9 \cdot 2 \cdot \sqrt[\large 3 \,]{10} = \;\)
\(\displaystyle 18 \cdot \sqrt[\large 3 \,]{10 \;} - 18 \cdot \sqrt[\large 3 \,]{10 \;} = \;\)\(\displaystyle \sqrt[\large 3 \,]{10 \;}(18 - 18) = \;\)\(\displaystyle 0\)
Cvičení 2.26 
Vyjádři ve tvaru jediné odmocniny za předpokladu, že \(a > 0\):
a) \(\displaystyle a^4 \cdot \sqrt{a^{- \, 2 \;} \cdot \sqrt[\large 3 \,]{a^{- \, 5 \;}}} =\)
\(\displaystyle a^4 \cdot \sqrt{\sqrt[\large 3 \,]{\left(a^{- \, 2}\right)^3 \;} \cdot \sqrt[\large 3 \,]{a^{- \, 5} \;}} = \;\)
\(\displaystyle a^4 \cdot \sqrt{\sqrt[\large 3 \,]{a^{- \, 6} \cdot a^{- \, 5} \;}} = \;\)\(\displaystyle a^4 \cdot \sqrt[\large 6 \,]{a^{- \, 11} \;} = \;\)
\(\displaystyle = \sqrt[\large 6 \,]{\left(a^{4}\right)^6 \;} \cdot \sqrt[\large 6 \,]{a^{- \, 11} \;} = \;\)
\(\displaystyle \sqrt[\large 6 \,]{a^{24} \cdot a^{- \, 11} \;} = \;\)\(\displaystyle \sqrt[\large 6 \,]{a^{13} \;}\)
b) \(\displaystyle a \cdot \sqrt{\frac {1}{a} \cdot \sqrt{\frac {1}{a^3} \;} \;} =\)
\(\displaystyle a \cdot \sqrt{\sqrt{\left(\frac {1} {a}\right)^2 \;} \cdot \sqrt{\frac {1} {a^3} \;} \;} = \;\)
\(\displaystyle a \cdot \sqrt{\sqrt{\frac {1}{a^2} \cdot \frac {1} {a^3} \;} \;} = \;\)\(\displaystyle a \cdot \sqrt{\sqrt{\frac {1} {a^5} \;}\;} = \;\)\(\displaystyle a \cdot \sqrt[\large 4 \,]{\frac {1} {a^5} \;} = \;\)
\(\displaystyle = \sqrt[\large 4 \,]{a^4 \;} \cdot \sqrt[\large 4 \,]{\frac {1} {a^5} \;} =\)\(\displaystyle \sqrt[\large 4 \,]{a^4 \cdot \frac {1} {a^5} \;} = \;\)\(\displaystyle \sqrt[\large 4 \,]{\frac {1} {a} \;}\)
Cvičení 2.27
Odmocni:
a) \(\displaystyle \left(\sqrt[\large 4 \,]{\frac {1} {3^7} \;} - \sqrt[\large 4 \,]{3 \;} \right) \cdot \sqrt[\large 4 \,]{27 \;} =\)
\(\displaystyle \sqrt[\large 4 \,]{\frac {1} {3^7}\;} \cdot \sqrt[\large 4 \,]{27 \;} -
\sqrt[\large 4 \,]{3 \;} \cdot \sqrt[\large 4 \,]{27} = \;\)
\(\displaystyle \sqrt[\large 4 \,]{\frac {1} {3^7} \cdot 3^3 \;} - \sqrt[\large 4 \,]{3 \cdot 3^3} = \;\)
\(\displaystyle = \sqrt[\large 4 \,]{\frac {1} {3^4} \;} - \sqrt[\large 4 \,]{3^4 \;} = \;\)
\(\displaystyle \sqrt[\large 4 \,]{\left(\frac {1} {3}\right)^4 \;} - 3 = \;\)\(\displaystyle \frac {1} {3} - 3 = \;\)\(\displaystyle \frac {1} {3} - \frac {9} {3} = \;\)\(\displaystyle - \, \frac {8} {3}\)
b) \(\displaystyle \sqrt[\large 3 \,]{12 \;} \cdot \left(\sqrt[\large 3 \,]{18 \;} + \sqrt[\large 3 \,]{4 \;} \right) -
2 \cdot \sqrt[\large 3 \,]{2 \;} \cdot \sqrt[\large 3 \,]{3 \;} - 2 \cdot \sqrt[\large 3 \,]{2 \;} \cdot \sqrt[\large 3 \,]{108 \;} =\)
\(\displaystyle = \sqrt[\large 3 \,]{12 \;} \cdot \sqrt[\large 3 \,]{18 \;} +
\sqrt[\large 3 \,]{12 \;} \cdot \sqrt[\large 3 \,]{4 \;} - 2 \cdot \sqrt[\large 3 \,]{6 \;} - 2 \cdot \sqrt[\large 3 \,]{216 \;}= \;\)
\(\displaystyle = \sqrt[\large 3 \,]{2^2 \cdot 3 \;} \cdot \sqrt[\large 3 \,]{2 \cdot 3^2 \;} +
\sqrt[\large 3 \,]{2^2 \cdot 3 \,} \cdot \sqrt[\large 3 \,]{2^2 \;} - 2 \cdot \sqrt[\large 3 \,]{6 \;} - 2 \cdot \sqrt[\large 3 \,]{6^3 \;} = \;\)
\(\displaystyle = \sqrt[\large 3 \,]{2^3 \cdot 3^3 \;} + \sqrt[\large 3 \,]{2^3 \cdot 2 \cdot 3 \;} -
2 \cdot \sqrt[\large 3 \,]{6 \;} - 2 \cdot 6 = \;\)
\(\displaystyle 2 \cdot 3 + 2 \cdot \sqrt[\large 3 \,]{6 \;} - 2 \cdot \sqrt[\large 3 \,]{6 \;} - 12 = \;\)\(\displaystyle 6 - 12 = \;\)\(\displaystyle - \, 6\)
c) \(\displaystyle \frac {\sqrt[\large 3 \,]{\sqrt[\large 4 \,]{25 \;} \cdot \sqrt[\large 6 \,]{125 \;} \cdot
\sqrt[\large 10 \,]{5^5 \;} \;} \cdot \sqrt[\large 4 \,]{25 \;}} {\sqrt[\large 4 \,]{625 \;}} =\)
\(\displaystyle \sqrt[\large 3 \,]{\sqrt[\large 4 \,]{5^2 \;} \cdot \sqrt[\large 6]{5^3 \;} \cdot
\sqrt[\large 10 \,]{5^5 \;}} \cdot \frac {\sqrt[\large 4 \,]{25 \;}} {\sqrt[\large 4 \,]{625 \;}} = \;\)
\(\displaystyle = \sqrt[\large 3 \,]{\sqrt[\large 2 \cdot 2 \,]{5^2 \;} \cdot \sqrt[\large 2 \cdot 3 \,]{5^3 \;}
\cdot \sqrt[\large 2 \cdot 5 \,]{5^5 \;}} \cdot \sqrt[\large 4 \,]{\frac {5^2} {5^4}\;} = \;\)
\(\displaystyle \sqrt[\large 3 \,]{\sqrt{5} \cdot \sqrt{5} \cdot \sqrt{5} \;} \cdot \sqrt[\large 4 \,]{\frac {1} {25} \;} = \;\)
\(\displaystyle \sqrt[\large 3 \,]{\left(\sqrt{5}\right)^3 \;} \cdot \sqrt[\large 4 \,]{\frac {1} {5^2} \;} = \;\)
\(\displaystyle = \sqrt{\sqrt[\large 3 \,]{5^3 \;} \;} \cdot \sqrt[\large 2 \cdot 2 \,]{\left(\frac {1} {5}\right)^2 \;} = \;\)\(\displaystyle \sqrt{5 \;} \cdot \sqrt{\frac {1} {5} \;} = \; \)\(\displaystyle \sqrt{5 \;} \cdot \frac {1} {\sqrt{5 \;} \;} = \;\)\(\displaystyle 1\)
