Odpowiedź:
Zadanie 17
[tex]a)(3ab {}^{2} ) {}^{3} = {3}^{3} \times {a}^{3} \times ( {b}^{2} ) {}^{3} = 27a {}^{3} {b}^{6} [/tex]
[tex]b)( \frac{1}{2} \: {a}^{2} {b}^{3}) {}^{4} = ( \frac{1}{2} ) {}^{4 } \times ( {a}^{2} ) {}^{4} \times ( {b}^{3} ) {}^{4} = \frac{1}{16} \: {a}^{8} {b}^{12} [/tex]
[tex]c)( \frac{2ab {}^{5} }{c {}^{2} } ) {}^{2} = \frac{2 {}^{2} \times a {}^{2} \times (b {}^{5} ) {}^{2} }{c {}^{4} } = \frac{4a {}^{2} b {}^{10} }{ {c}^{4} } [/tex]
[tex]d)( \frac{a {}^{5}b {}^{6} }{3x {}^{2} } ) {}^{5} = \frac{(a {}^{5}) {}^{5} \times (b {}^{6}) {}^{5} }{ {3}^{5} \times ( {x}^{2}) {}^{5} } = \frac{ {a}^{25} {b}^{30} }{243 {x}^{10} } [/tex]
Pozdrawiam.
