Pliisssss.,,bantu jawab

A. Karena jika didalam suatu akar terdapat bilangan berpangkat 2 maka bilangan tersebut daoat diletakkan diluar)

[tex]a.) \: \sqrt{8} = 2 \sqrt{2} \\ \sqrt{4 \times 2} = 2 \sqrt{2} \\ 2 \sqrt{2} = 2 \sqrt{2} \\ \\ b.) \sqrt[3]{16} = 2\sqrt[3]{2} \\ \sqrt[3]{8 \times 2} = 2 \sqrt[3]{2} \\ 2 \sqrt[3]{2} = 2 \sqrt[3]{2} \\ \\ c.) \sqrt[4]{243} = 3 \sqrt[4]{3} \\ \sqrt[4]{81 \times 3} = 3 \sqrt[4]{3} \\ 3 \sqrt[4]{3} = 3 \sqrt[4]{3} \\ \\ d.) \sqrt[3]{81 {p}^{2} {q}^{4} {r}^{5} } = 3qr \sqrt[3]{3 {p}^{2}q {r}^{2} } \\ \sqrt[3]{27 \times 3 {p}^{2} {q}^{4} {r}^{5 } } = 3qr \sqrt[3]{3 {p}^{2}q {r}^{2} } \\ 3qr \sqrt[3]{3 {p}^{2}q {r}^{2} } = 3qr \sqrt[3]{3 {p}^{2}q {r}^{2} } \\ \\ e.) \sqrt{ {a}^{5} } = {a}^{2} \sqrt{a} \\ \sqrt{a \times {a}^{4} } = {a}^{2} \sqrt{a} \\ {a}^{2} \sqrt{a} = {a}^{2} \sqrt{a}\\ \\ f.) \sqrt[3]{ {x}^{4} {y}^{3} } = xy \sqrt[3]{x} \\ \sqrt[3]{ {x}^{3} \times x \times {y}^{3} } = xy \sqrt[3]{x} \\ xy \sqrt[3]{x} = xy \sqrt[3]{x} \\ \\ semoga \: membantu[/tex]