Question

Tolong bantuin dong tugas mtk ku​

Answer 1

Jawab:
[tex]\displaystyle\rm(3a)\:\:\:\bf=\frac{\sqrt{21}}{7}\\\\\rm(3b)\:\:\:\bf=2\sqrt{7}-2\sqrt{5}\\\\\rm(3c)\:\:\:\bf=1+\sqrt{6}[/tex]

Penjelasan:
Sifat pecahan...
[tex]\displaystyle\frac{a}{b}=\frac{ab}{b^2}\\\\\frac{a}{b+c}=\frac{a(b-c)}{b^2-c^2}\\\\\frac{a}{b-c}=\frac{a(b+c)}{b^2-c^2}[/tex]
Sifat pengakaran
[tex](\sqrt{p})^2=p[/tex]

Rasionalkanlah!
[tex]\displaystyle\rm(3a)\:\:\:\frac{3}{\sqrt{21}}=\frac{3\sqrt{21}}{(\sqrt{21})^2}=\frac{3\sqrt{21}}{21}\bf=\frac{\sqrt{21}}{7}\\\\\rm(3b)\:\:\:\frac{4}{\sqrt{7}+\sqrt{5}}=\frac{4(\sqrt{7}-\sqrt{5})}{(\sqrt{7})^2-(\sqrt{5})^2}\\\\=\frac{4(\sqrt{7}-\sqrt{5})}{7-5}=\frac{4(\sqrt{7}-\sqrt{5})}{2}=2(\sqrt{7}-\sqrt{5})\\\\\bf=2\sqrt{7}-2\sqrt{5}[/tex]

[tex]\displaystyle\rm(3c)\:\:\frac{\sqrt{8}-\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{8}-\sqrt{3})(\sqrt{3}+\sqrt{2})}{(\sqrt{3})^2-(\sqrt{2})^2}\\\\=\frac{(\sqrt{4\cdot2}-\sqrt{3})(\sqrt{3}+\sqrt{2})}{3-2}\\\\=\frac{((\sqrt{4}\sqrt{2})-\sqrt{3})(\sqrt{3}+\sqrt{2})}{1}\\\\=\frac{(2\sqrt{2}-\sqrt{3})(\sqrt{3}+\sqrt{2})}{1}\\\\=(2\sqrt{2}-\sqrt{3})(\sqrt{3}+\sqrt{2})\\\\=2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{2}-\sqrt{3}\sqrt{3}-\sqrt{3}\sqrt{2}[/tex]

[tex]\displaystyle=2\sqrt{2(3)}+2(\sqrt{2})^2-(\sqrt{3})^2-\sqrt{3(2)}\\\\=2\sqrt{6}+2(2)-3-\sqrt{6}\\\\=2\sqrt{6}+4-3-\sqrt{6}\\\\=\bf1+\sqrt{6}[/tex]

(xcvi)