Question

√4+∛8+√25=
√64-√49+√1=
√5+√16=
√7,5+1=
∛125+√16-√18=
∛√25+√9=
5√19-∛8=
√10+√36=
∛27+4√14-∛32=
√8+∛1=

Answer 1

Tu tienes definición:
$\sqrt{a}=b \qquad \Rightarrow \qquad b^{2}=a \qquad a,b \ge 0 \\ \sqrt[3]{a}=b \qquad \Rightarrow \qquad b^{3}=a$
$\sqrt{4}+\sqrt[3]{8}+\sqrt{25}=2+2+5=9 \\ \sqrt{64}-\sqrt{49}+\sqrt{1}=8-7+1=2 \\ \sqrt{5}+\sqrt{16}=\sqrt{5}+4 \\ \sqrt{7,5}+1 \ \hbox{es no contar} \\ \sqrt[3]{125}+\sqrt{16}-\sqrt{18}=5+4-\sqrt{9 \cdot 2}=9-3\sqrt{2} \\ \sqrt[3]{\sqrt{25}+\sqrt{9}}=\sqrt[3]{5+3}=\sqrt[3]{8}=2 \\ 5\sqrt{19}-\sqrt[3]{8}=5\sqrt{19}-2 \\ \sqrt{10}+\sqrt{36}=\sqrt{10}+6 \\ \sqrt[3]{27}+4\sqrt{14}-\sqrt[3]{32}=3+4\sqrt{14}-\sqrt[3]{8 \cdot 4}=3+4\sqrt{14}-2\sqrt[3]{4} \\ \sqrt{8}+\sqrt[3]{1}=\sqrt{4 \cdot 2}+1=2\sqrt{2}+1$