Question

separa en terminos y resuelve



$\sqrt{121} * 5 +(4+2)^{2} =\\\\127* 42- 9^{2} *(7-2)+ 48=\\\\2345*\sqrt{25} +282/2^{1 }-\sqrt[3]{27} \\\\(\frac{3}{4} *\frac{1}{2} )/\frac{3}{8} +3\frac{1}{12} =\\\\18* 26+3^{3} * 2^{5} -} \sqrt[3]{125=}$

POR FAVOR AYADA

Answer 1

Respuesta:

Explicación paso a paso:

$\sqrt{121} *5+(4+2)^{2} =$

 $11 * 5 +(6)^{2}$

$55 + 36 = 91$

$127*42-9^{2} *(7-2)+48=$

$5334 - 81 * 5 + 48 =$

$5334 - 405 + 48 = 4977$

$2345 * \sqrt{25} +282/2^{1} -\sqrt[3]{27} =$

$2345 * 5 + 282/2 - 3 =$

$11725 + 141 - 3 = 11863$

$(\frac{3}{4} *\frac{1}{2} )/\frac{3}{8} +3\frac{1}{12} =$

$\frac{1}{2} /\frac{3}{8} +\frac{37}{12} =$

$\frac{8}{6} +\frac{37}{12}$

$mcm = 12$

$\frac{16 + 37}{12} = \frac{53}{12}$

$18*26+3^{3} *2^{5} -\sqrt[3]{125} =$

$468 + 27 * 32 - 5 =$

$468 + 864 - 5 = 1327$