Bantu jawab nomor 4-6 kak,

Mapel : Matematika
Materi : Akar Pangkat

Soal No. 4

[tex]= \frac{2 \sqrt{3}-5}{3- \sqrt{2}}\\\\=\frac{2 \sqrt{3}-5}{3- \sqrt{2}}\times\frac{3+ \sqrt{2}}{3+ \sqrt{2}}\\\\=\frac{(2 \sqrt{3}-5)(3+ \sqrt{2})}{3^2- (\sqrt{2})^2}\\\\=\frac{(2 \sqrt{3})(3)-(5)(3)+(2 \sqrt{3})(\sqrt{2})-(5)(\sqrt{2})}{9- 2}\\\\=\frac{(6 \sqrt{3})-(15)+(2 \sqrt{6})-(5\sqrt{2})}{7}[/tex]

Soal No. 5

[tex]= \frac{5-2 \sqrt{2}}{2\sqrt{2}-5}\\\\=\frac{5-2 \sqrt{2}}{2\sqrt{2}-5}\times\frac{2\sqrt{2}+5}{2\sqrt{2}+5}\\\\=\frac{(5-2 \sqrt{2})(2\sqrt{2}+5)}{(2\sqrt{2})^2-(5)^2}\\\\=\frac{(5)(2\sqrt{2})-(2\sqrt{2})(2\sqrt{2})+(5)(5)-(2 \sqrt{2})(5)}{8-25}\\\\=\frac{(10\sqrt{2})-(8)+(25)-(10\sqrt{2})}{-17} \\ \\ =\frac{17}{-17} \\ \\ =-1 [/tex]

Soal No. 6

[tex]= \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}+\sqrt{3}}\\\\= \frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}+\sqrt{3}}\times\frac{\sqrt{6}-\sqrt{3}}{\sqrt{6}-\sqrt{3}}\\\\= \frac{(\sqrt{6}-\sqrt{3})(\sqrt{6}-\sqrt{3})}{(\sqrt{6}+\sqrt{3})(\sqrt{6}-\sqrt{3})}\\\\= \frac{(\sqrt{6})^2-(\sqrt{3})(\sqrt{6})-(\sqrt{6})(\sqrt{3})+(\sqrt{3})^2}{(\sqrt{6})^2-(\sqrt{3})^2}\\\\=\frac{6-2(\sqrt{18})+3}{6-3}\\\\=\frac{9-2(\sqrt{9\times2})}{3}\\\\=\frac{9-6\sqrt{2}}{3} \\ \\ =3-2 \sqrt{2} [/tex]

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