Question

Ayúdenme a resolver estos ejercicios por favor ​

Answer 1

$\sqrt[3]{1}+\sqrt{4} =1+\sqrt{4}=1+2=3\\\\ 4^{\frac{3}{2} } =\sqrt{4^3} =\sqrt{64} =8\\\\2*4^{\frac{1}{2} }=2*\sqrt{4} =2*2=4\\\\\sqrt{3}*\sqrt{3^{2} } *\sqrt[5]{7^{5} }= \sqrt{3}*3*7=21\sqrt{3} \\\\4\sqrt{9}+5\sqrt{9} =9\sqrt{9}=9*3=27\\\\4\sqrt{2}*\sqrt{8} =4\sqrt{2} *\sqrt{4*2} =4\sqrt{2} *\sqrt{4}\sqrt{2} =4\sqrt{2} *2\sqrt{2} =8*2^{\frac{2}{2} }=8*2=16\\\\(2\sqrt[3]{2^2} ) ^3=(2*2^\frac{2}{3} )^3=(2^\frac{5}{3})^3 =2^\frac{15}{3}=2^5=32$

$\sqrt{144}*\sqrt{2^4} =12*2^\frac{4}{2} =12*2^2=12*4=48$