▶️ Ejercicio 205 álgebra - Solucionario Baldor ✅✅

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CAPITULO XXVIII

Problemas sobre ecuaciones simultaneas

Ejercicio 205

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${(4 a^{2})}^{2} = 4^{2} a^{2 \times 2} = 16 a^{4}$
${(- 5 a)}^{3} = {(- 5)}^{3} a^{3} = - 125 a^{3}$
${(3 x y)}^{3} = 3^{3} x^{3} y^{3} = 27 x^{3} y^{3}$
${(- 6 a^{2} b)}^{2} = {(- 6)}^{2} a^{2 \times 2} b^{2} = 36 a^{4} b^{2}$
${(- 2 x^{2} y^{3})}^{3} = {(- 2)}^{3} x^{2 \times 3} y^{3 \times 3} = - 8 x^{6} y^{9}$
${(4 a^{2} b^{3} c^{4})}^{3} = 4^{3} a^{2 \times 3} b^{3 \times 3} c^{4 \times 3} = 64 a^{6} b^{9} c^{12}$
${(- 6 x^{4} y^{5})}^{2} = {(- 6)}^{2} x^{4 \times 2} y^{5 \times 2} = 36 x^{8} y^{10}$
${(- 7 a b^{3} c^{4})}^{3} = {(- 7)}^{3} a^{3} b^{3 \times 3} c^{4 \times 3} = - 343 a^{3} b^{9} c^{12}$
${(a^{m} b^{n})}^{x} = a^{m x} b^{n x}$
${(- 2 x^{3} y^{5} z^{6})}^{4} = {(- 2)}^{4} x^{3 \times 4} y^{5 \times 4} z^{6 \times 4} = 16 x^{12} y^{20} z^{24}$
${(- 3 m^{3} n)}^{3} = {(- 3)}^{3} m^{3 \times 3} n^{3} = - 27 m^{9} n^{3}$
${(a^{2} b^{3} c)}^{m} = a^{2 m} b^{3 m} c^{m}$
${(- m^{2} n x^{3})}^{4} = {(- m^{2})}^{4} n^{4} x^{3 \times 4} = m^{8} n^{4} x^{12}$
${(- 3 a^{2} b)}^{5} = {(- 3)}^{5} a^{2 \times 5} b^{5} = - 243 a^{10} b^{5}$
${(7 x^{5} y^{6} z^{8})}^{2} = 7^{2} x^{5 \times 2} y^{6 \times 2} z^{8 \times 2} = 49 x^{10} y^{12} z^{16}$
${(- \frac{x}{2 y})}^{2} = \frac{{(- x)}^{2}}{2^{2} y^{2}} = \frac{x^{2}}{4 y^{2}}$
${(- \frac{2 m}{n^{2}})}^{3} = \frac{{(- 2 m)}^{3}}{n^{2 \times 3}} = - \frac{8 m^{3}}{n^{6}}$
${(\frac{a b^{2}}{5})}^{3} = \frac{a^{3} b^{2 \times 3}}{5^{3}} = \frac{a^{3} b^{6}}{125}$
${(- \frac{3 x^{2}}{4 y})}^{2} = \frac{{(- 3)}^{2} x^{2 \times 2}}{4^{2} y^{2}} = \frac{9 x^{4}}{16 y^{2}}$
${(- \frac{2 a b^{2}}{3 m^{3}})}^{4} = \frac{{(- 2)}^{4} a^{4} b^{2 \times 4}}{3^{4} m^{3 \times 4}} = \frac{16 a^{4} b^{8}}{81 m^{12}}$
${(\frac{2 m^{3} n}{3 x^{4}})}^{5} = \frac{2^{5} m^{3 \times 5} n^{5}}{3^{5} x^{4 \times 5}} = \frac{32 m^{15} n^{5}}{243 x^{20}}$
${(- \frac{3}{4} a^{3} b^{2})}^{2} = {(- \frac{3}{4})}^{2} a^{3 \times 2} b^{2 \times 2} = \frac{9}{16} a^{6} b^{4}$
${(- \frac{1}{3} m n^{2})}^{4} = {(- \frac{1}{3})}^{4} m^{4} n^{2 \times 4} = \frac{1}{81} m^{4} n^{8}$
${(- \frac{1}{2} a^{2} b^{4})}^{5} = {(- \frac{1}{2})}^{5} a^{2 \times 5} b^{4 \times 5} = - \frac{1}{32} a^{10} b^{20}$