a. 8 akar 10 per akar 5
b.12 3+akar 5
c.10 akar 2 per akar 7 - akar 5
tolong yaa kak makasih
Jawab:
[tex]\rm (a.)~~\frac{8\sqrt{10}}{\sqrt{5}}=\bf8\sqrt{2}\\\\\rm(b.)~~\frac{12}{3+\sqrt{5}}=\bf9-3\sqrt{5}\\\\\rm(c.)~~\frac{10\sqrt{2}}{\sqrt{7}-\sqrt{5}}=\bf5\sqrt{14}+5\sqrt{10}[/tex]
Penjelasan dengan langkah-langkah:
[tex]\rm (a.)~~\frac{8\sqrt{10}}{\sqrt{5}}=\frac{8\sqrt{10}\sqrt{5}}{\sqrt{5}\sqrt{5}}\\=\frac{8\sqrt{2\cdot5}\sqrt{5}}{5}=\frac{8\sqrt{2}\sqrt{5}\sqrt{5}}{5}\\=\frac{8\sqrt{2}\cdot\not5}{\not5}\\=\bf8\sqrt{2}\\\\\rm(b.)~~\frac{12}{3+\sqrt{5}}=\frac{12}{3+\sqrt{5}}\cdot\frac{3-\sqrt{5}}{3-\sqrt{5}}\\=\frac{12(3-\sqrt{5})}{3^2-\sqrt{5}^2}=\frac{12(3-\sqrt{5})}{9-5}=\frac{12(3-\sqrt{5})}{4}\\=\frac{12}{4}(3-\sqrt{5})=3(3-\sqrt{5})\\=3(3)-3\sqrt{5}\\=\bf9-3\sqrt{5}[/tex]
[tex]\rm(c.)~~\frac{10\sqrt{2}}{\sqrt{7}-\sqrt{5}}=\frac{10\sqrt{2}}{\sqrt{7}-\sqrt{5}}\cdot\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\=\frac{10\sqrt{2}(\sqrt{7}+\sqrt{5})}{\sqrt{7}^2-\sqrt{5}^2}=\frac{10\sqrt{2}(\sqrt{7}+\sqrt{5})}{7-5}\\=\frac{10\sqrt{2}(\sqrt{7}+\sqrt{5})}{2}=\frac{10}{2}(\sqrt{2}(\sqrt{7}+\sqrt{5}))\\=5\sqrt{2}(\sqrt{7}+\sqrt{5})=5\sqrt{2\cdot7}+5\sqrt{2\cdot5}\\=\bf5\sqrt{14}+5\sqrt{10}[/tex]
(xcvi)
[answer.2.content]
