Question

Resolver las siguientes Expresiones Algebraicas ​

Answer 1

Verified answer

1.

[tex]2{c}^{3} \sqrt{16ac} = 2{(4)}^{3} \sqrt{16(1)(4)} = 128 \sqrt{64} = 1024[/tex]

2.

[tex] \frac{3{(2)}^{3} {( \frac{1}{3} )}^{2}}{(1)(2)( \frac{1}{3})} = \frac{ \frac{8}{3} }{ \frac{2}{3} } = 4[/tex]

3.

[tex] \frac{4 {( \frac{3}{2})}^{2} ( \frac{1}{5})}{6 \sqrt[3]{30 + 2} } = \frac{ \frac{9}{5} }{6 \sqrt[3]{32} } = \frac{ \frac{9}{5} }{12 \sqrt[3]{4} } = \frac{108 \sqrt[3]{4}}{5} [/tex]

4.

[tex] \frac{ {4}^{3} }{5} - \frac{ {1}^{2} }{2} + \frac{{(\frac{3}{2})}^{2} }{3} = \frac{64}{5} - \frac{1}{2} + \frac{3}{4} [/tex]

[tex] = \frac{256 - 10 + 15}{20} = \frac{261}{20}[/tex]

5.

[tex] \frac{ \frac{3}{5} - 4 }{4} + \frac{6 - \frac{27}{8} }{ \frac{1}{3} } - \frac{ \frac{1}{5} }{2} = - \frac{17}{20} + \frac{ \frac{21}{8} }{ \frac{1}{3} } - \frac{1}{10} [/tex]

= [tex] \frac{63}{8} - \frac{19}{20} = \frac{315 - 38}{40} = \frac{277}{40} [/tex]