\begin{align}
\end{align}
Cvičení - Součin, podíl a mocniny lomených výrazů
Cvičení 4.9
Rozhodni, zda pro libovolné výrazy \(V_1\), \(V_2\),
\(V_3\), \(V_4\), přičemž pro všechny hodnoty proměnných je
\(V_1 \neq 0\), \(V_2 \neq 0\), \(V_3 \neq 0\),
\(V_4 \neq 0\) a \(V_1 \neq V_2\), platí:
Cvičení 4.10
Vypočítej:
a) \(\displaystyle \frac {4ab^2} {a + b} \cdot \frac {a^2 + 2ab + b^2} {8a^2b} = \;\)
\(\displaystyle \frac {4ab^2} {a + b} \cdot \frac {(a + b)^2} {8a^2b} = \;\)\(\displaystyle \frac {b} {1} \cdot \frac {a + b} {2a} = \;\)\(\displaystyle \frac {b \cdot (a + b)} {1 \cdot 2a} = \;\)\(\displaystyle \frac {b(a + b)} {2a}\)
Podmínky: \(a \in \mathbb R - \{0\}\), \(b \in \mathbb R - \{0\}\)
a zároveň \(a \neq - \, b\)
b) \(\displaystyle \frac {5a^2 + 5a} {9 + 3a} \cdot \frac {a^2 - 9} {5a - 10} \cdot \frac {6a - 12} {2a^2 + 2a} = \;\)
\(\displaystyle \frac {5a(a + 1)} {3(3 + a)} \cdot \frac {(a + 3)(a - 3)} {5(a - 2)} \cdot \frac {6(a - 2)} {2a(a + 1)} = \;\)
\(\displaystyle = 1 \cdot \frac {a - 3} {1} \cdot 1 = \;\)\(\displaystyle a - 3\)
Podmínky: \(a \in \mathbb R - \{-\,3; -\,1; 0; 2\}\)
c) \(\displaystyle \frac {2a^2 - 4ab + 2b^2} {(8a^2 - 8b^2)(a - b)} \cdot (4a + 4b) = \;\)
\(\displaystyle \frac {2(a^2 - 2ab + b^2)} {8(a^2 - b^2)(a - b)} \cdot \frac {4(a + b)} {1} = \;\)
\(\displaystyle = \frac {2(a - b)^2} {8(a + b)(a - b)(a - b)} \cdot \frac {4(a + b)} {1} = \;\)\(1\)
Podmínky: \(a \in \mathbb R\), \(b \in \mathbb R\)
a zároveň \(a \neq \pm b\)
d) \(\displaystyle 7a \cdot \frac {16 - a^2} {21a^2 + 14a} \cdot \frac {9a + 6} {a - 4} = \;\)
\(\displaystyle \frac {7a} {1} \cdot \frac {(4 - a)(4 + a)} {7a(3a + 2)} \cdot \frac {3(3a + 2)} {a - 4} = \;\)
\(\displaystyle = 1 \cdot \frac {(4 - a)(4 + a)} {1} \cdot \frac {3} {(-1) \cdot (4 - a)} = \;\)\(\displaystyle \frac {(4 + a) \cdot 3} {-1} = \; \)\(\displaystyle -12 - 3a\)
Podmínky: \(\displaystyle a \in \mathbb R - \left\{- \, \frac {2} {3}; 0; 4\right\}\)
Cvičení 4.11
Cvičení 4.12
Vypočítej:
a) \(\displaystyle \left(\frac {4p^2} {3q}\right)^3 \cdot \left(\frac {9q} {2p}\right)^2 = \;\)
\(\displaystyle \frac {\left(4p^2\right)^3} {(3q)^3} \cdot \frac {(9q)^2} {(2p)^2} = \;\)
\(\displaystyle \frac {4^{1 \cdot 3}p^{2 \cdot 3}} {3^3q^3} \cdot \frac {9^2q^2} {2^2p^2} = \;\)
\(\displaystyle \frac {64p^6} {27q^3} \cdot \frac {81q^2} {4p^2} = \;\)\(\displaystyle \frac {16p^4} {q} \cdot \frac {3} {1} = \;\)\(\displaystyle \frac {48p^4} {q}\)
Podmínky: \(p \in \mathbb R - \{0\}\), \(q \in \mathbb R - \{0\}\)
b) \(\displaystyle \left(\frac {p^2q - q} {p + 1}\right)^4 \cdot \left(\frac {15} {5pq^2 - 5q^2}\right)^3 = \;\)
\(\displaystyle \left(\frac {q(p^2 - 1)} {p + 1}\right)^4 \cdot \left(\frac {15} {5q^2(p - 1)}\right)^3 = \;\)
\(\displaystyle = \left(\frac {q(p + 1)(p - 1)} {p + 1}\right)^4 \cdot \left(\frac {3} {q^2(p - 1)}\right)^3 = \;\)
\(\displaystyle \left(\frac {q(p - 1)} {1}\right)^4 \cdot \frac {3^3} {q^6(p - 1)^3} = \;\)
\(\displaystyle \frac {q^4(p - 1)^4} {1} \cdot \frac {27} {q^6(p - 1)^3} = \;\)
\(\displaystyle = \; \frac {p - 1} {1} \cdot \frac {27} {q^2} = \;\)
\(\displaystyle \frac {27(p - 1)} {q^2}\)
Podmínky: \(p \in \mathbb R - \{\pm 1\}\), \(q \in \mathbb R - \{0\}\)
c) \(\displaystyle \left(\frac {2p + 3} {p - 5}\right)^{-2} \cdot \frac {4p^2 + 12p + 9} {25 - 5p} = \;\)
\(\displaystyle \frac {(p - 5)^2} {(2p + 3)^2} \cdot \frac {(2p + 3)^2} {5(5 - p)} = \;\)
\(\displaystyle \frac {(p - 5)^2} {1} \cdot \frac {1} {5 \cdot (-1) \cdot (p - 5)} = \;\)
\(\displaystyle = \; \frac {p - 5} {- \, 5} = \;\)
\(\displaystyle \frac {5 - p} {5}\)
Podmínky: \(\displaystyle p \in \mathbb R - \left\{- \, \frac {3} {2}; 5\right\}\)
d) \(\displaystyle \frac {(p^3 + 27)^3} {(p - 3)^2} \cdot \left(p + 3\right)^{-3} = \;\)
\(\displaystyle \frac {\left[(p + 3)(p^2 - 3p + 9)\right]^3} {(p - 3)^2} \cdot \left(\frac {p + 3} {1}\right)^{-3} = \;\)
\(\displaystyle = \frac {(p + 3)^3(p^2 - 3p + 9)^3} {(p - 3)^2} \cdot \frac {1} {(p + 3)^3} = \;\)
\(\displaystyle \frac {(p^2 - 3p + 9)^3} {(p - 3)^2}\)
Podmínky: \(p \in \mathbb R - \{\pm 3\}\)
Cvičení 4.13
Rozhodni, zda pro libovolné výrazy \(V_1\), \(V_2\),
\(V_3\), \(V_4\), přičemž pro všechny hodnoty proměnných je
\(V_2 \neq 0\), \(V_3 \neq 0\),
\(V_4 \neq 0\) a \(V_1 \neq V_2\), platí:
Cvičení 4.14
Vypočítej:
a) \(\displaystyle \frac {49t^2 - 1} {14t^3 + 2t^2} \div \frac {1 - 7t} {14t^2} = \;\)
\(\displaystyle \frac {49t^2 - 1} {14t^3 + 2t^2} \cdot \frac {14t^2} {1 - 7t} = \;\)
\(\displaystyle \frac {(7t - 1)(7t + 1)} {2t^2(7t + 1)} \cdot \frac {14t^2} {(-1)(7t - 1)} = \;\)
\(\displaystyle 1 \cdot \frac {7} {(-1)} = \;\)
\(\displaystyle - \, 7\)
Podmínky: \(\displaystyle t \in \mathbb R - \left\{- \, \frac {1} {7}; 0; \frac {1} {7}\right\}\)
b) \(\displaystyle \frac {tu - 2t + 5u - 10} {2 - u} \div (t^2 + 10t + 25)^2 = \;\)
\(\displaystyle \frac {tu - 2t + 5u - 10} {2 - u} \cdot \frac {1} {(t^2 + 10t + 25)^2} = \;\)
\(\displaystyle = \frac {t(u - 2) + 5(u - 2)} {2 - u} \cdot \frac {1} {\left[(t + 5)^2\right]^2} = \;\)
\(\displaystyle \frac {(u - 2)(t + 5)} {(-1)(u - 2)} \cdot \frac {1} {(t + 5)^4} = \;\)
\(\displaystyle \frac {1} {- \, 1} \cdot \frac {1} {(t + 5)^3} = \;\)
\(\displaystyle \frac {- \, 1} {(t + 5)^3}\)
Podmínky: \(t \in \mathbb R - \{- \, 5\}\), \(u \in \mathbb R - \{2\}\)
c) \(\displaystyle \LARGE \frac {\; \; \; \; \; \; \Large \frac {8t^3 \, - \, 125} {(t \, - \, 1)^3}\; \; \; \; \; \;}
{\Large \frac {2t \, - \, 5} {t^3 \, - \, 3t^2 \, + \, 3t \, - \, 1}} \normalsize = \;\)
\(\displaystyle \frac {8t^3 - 125} {(t - 1)^3} \div \frac {2t - 5} {t^3 - 3t^2 + 3t - 1} = \;\)
\(\displaystyle \frac {(2t)^3 - 5^3} {(t - 1)^3} \cdot \frac {t^3 - 3t^2 + 3t - 1} {2t - 5} = \;\)
\(\displaystyle = \frac {(2t - 5)(4t^2 + 10t + 25)} {(t - 1)^3} \cdot \frac {(t - 1)^3} {2t - 5} = \;\)
\(\displaystyle \frac {4t^2 + 10t + 25} {1} \cdot 1 = \;\)
\(\displaystyle 4t^2 + 10t + 25\)
Podmínky: \(\displaystyle t \in \mathbb R - \left\{1; \frac {5} {2}\right\}\)
d) \(\displaystyle \LARGE \frac {\; \; \; \; \Large \frac {6t^2 \, + \, 6tu} {7u \, - \, 7u^2}\; \; \; \;}
{\Large \frac {18u^2 \, + \, 18tu} {21t \, - \, 21tu}} \normalsize \cdot \left(\frac {t} {u}\right)^{-2} = \;\)
\(\displaystyle \left(\frac {6t^2 + 6tu} {7u - 7u^2} \div \frac {18u^2 + 18tu} {21t - 21tu}\right)
\cdot \frac {u^2} {t^2} = \;\)
\(\displaystyle \frac {6t(t + u)} {7u(1 - u)} \cdot \frac {21t(1 - u)} {18u(u + t)} \cdot \frac {u^2} {t^2} = \;\)
\(\displaystyle 1\)
Podmínky: \(t \in \mathbb R - \{0\}\), \(u \in \mathbb R - \{0; 1\}\)
a zároveň \(t \neq - \, u\)
Cvičení 4.15 
Vypočítej:
a) \(\displaystyle \LARGE \frac {\; \; \Large \frac {9u^2 \, + \, 24uv \, + \, 16v^2} {4u^2 \, - \, 25v^2}\; \;}
{\Large \frac {24u \, + \, 32v} {2u \, - \, 5v}} \cdot
\LARGE \frac {\Large \frac {(2u \, - \, 2v)^3} {6u^2 \, + \, 8uv \, + \, 20v^2 \, + \, 15uv}}
{\; \; \Large \frac {u^3 \, - \, 3u^2v \, + \, 3uv^2 \, - \, v^3} {8u^3 \, + \, 60u^2v \, + \, 150uv^2 \, + \, 125v^3}\; \;} \normalsize = \)
\(\displaystyle = \left(\frac {9u^2 + 24uv + 16v^2} {4u^2 - 25v^2} \div \frac {24u + 32v} {2u - 5v}\right) \cdot
\left(\frac {(2u - 2v)^3} {6u^2 + 8uv + 20v^2 + 15uv} \div
\frac {u^3 - 3u^2v + 3uv^2 - v^3} {8u^3 + 60u^2v + 150uv^2 + 125v^3} \right) =\)
\(\displaystyle = \frac {(3u + 4v)^2} {(2u + 5v)(2u - 5v)} \cdot \frac {2u - 5v} {8(3u + 4v)} \cdot
\frac {2^3(u - v)^3} {2u(3u + 4v) + 5v(4v + 3u)} \cdot \frac {8u^3 + 60u^2v + 150uv^2 + 125v^3} {u^3 - 3u^2v + 3uv^2 - v^3} =\)
\(\displaystyle = \frac {3u + 4v} {2u + 5v} \cdot \frac {1} {8} \cdot
\frac {8(u - v)^3} {(3u + 4v)(2u + 5v)} \cdot \frac {(2u + 5v)^3} {(u - v)^3} = \;\)
\(\displaystyle 2u + 5v\)
Podmínky: \(u \in \mathbb R\), \(v \in \mathbb R\) a zároveň
\(\displaystyle u \neq \pm \frac {5} {2}v \wedge u \neq - \,\frac {4} {3}v \wedge u \neq v\)
b) \(\displaystyle \LARGE \frac {\; \; \Large \frac {64u^3 \, + \, 125} {12u \, + \, 15 \, + \, 8uv \, + \, 10v}\; \;}
{\Large \frac {8u \, + \, 8v} {128}} \large \div
\LARGE \frac {\; \; \Large \frac {3u \, + \, 2uv \, - \, 3v \, - \, 2v^2} {8v^3 \, + \, 36v^2 \, + \, 54v \, + \, 27}\; \;}
{\Large \frac {u^2 \, - \, v^2} {48 \, + \, 32v}} \normalsize =\)
\(\displaystyle = \left(\frac {64u^3 + 125} {12u + 15 + 8uv + 10v} \div \frac {8u + 8v} {128}\right) \div
\left(\frac {3u + 2uv - 3v - 2v^2} {8v^3 + 36v^2 + 54v + 27} \div \frac {u^2 - v^2} {48 + 32v} \right) =\)
\(\displaystyle = \left(\frac {(4u)^3 + 5^3} {3(4u + 5) + 2v(4u + 5)} \cdot \frac {128} {8(u + v)}\right) \div
\left(\frac {u(3 + 2v) - v(3 + 2v)} {(2v + 3)^3} \cdot \frac {16(3 + 2v)} {(u + v)(u - v)} \right) =\)
\(\displaystyle = \left(\frac {(4u + 5)(16u^2 - 20u + 25)} {(4u + 5)(3 + 2v)} \cdot \frac {16} {u + v}\right) \div
\left(\frac {(3 + 2v)(u - v)} {(2v + 3)^3} \cdot \frac {16(3 + 2v)} {(u + v)(u - v)}\right) =\)
\(\displaystyle = \frac {(4u + 5)(16u^2 - 20u + 25)} {(4u + 5)(3 + 2v)} \cdot \frac {16} {u + v} \cdot
\frac {(2v + 3)^3} {(3 + 2v)(u - v)} \cdot \frac {(u - v)(u + v)} {16(3 + 2v)} = \;\)
\(\displaystyle 16u^2 - 20u + 25\)
Podmínky: \(\displaystyle u \in \mathbb R - \left\{- \, \frac {5} {4}\right\}\),
\(\displaystyle v \in \mathbb R - \left\{- \, \frac {3} {2}\right\}\) a zároveň
\(u \neq \pm v\)
