Tolong bantuannya sama caranya plissss no 24&25

——23——
Mapel : Matematika
Bab : Akar dan Pangkat

[tex] \frac{2 + \sqrt{8} }{ \sqrt{6} } \\ = \frac{2 + \sqrt{8} }{ \sqrt{6} } \times \frac{ \sqrt{6} }{ \sqrt{6} } \\ = \frac{(2 + \sqrt{8}) \times \sqrt{6} }{6} \\ = \frac{(2 + \sqrt{(4 \times 2)} ) \times \sqrt{6}}{6} \\ = \frac{(2 + 2 \sqrt{2} ) \times \sqrt{6} }{6} \\ = \frac{2(1 + \sqrt{2}) \times \sqrt{6} }{6} \\ = \frac{(1 + \sqrt{2}) \times \sqrt{6} }{3} \\ = \frac{ \sqrt{6} + \sqrt{12} }{3} \\ = \frac{1}{ 3} \times ( \sqrt{6} + \sqrt{12} ) \\ = \frac{1}{3} \times ( \sqrt{6} + \sqrt{(4 \times 3)} \\ = \frac{1}{3} \times ( \sqrt{6} + 2 \sqrt{3} )[/tex]

Jawaban : A. ⅓ × (√6 + 2√3)

——24——
Mapel : Matematika
Bab : Akar dan Pangkat

[tex]m \times (2 \sqrt{3} - \sqrt{7}) = 10 \\ m =10 \div (2 \sqrt{3} - \sqrt{7} ) \\ m = \frac{10}{2 \sqrt{3} - \sqrt{7} } \times \frac{2 \sqrt{3} + \sqrt{7} }{2 \sqrt{3} + \sqrt{7} } \\ m = \frac{10 \times (2 \sqrt{3} + \sqrt{7}) }{(4 \times 3) - 7} \\ m = \frac{10 \times (2 \sqrt{3} + \sqrt{7}) }{12 - 7} \\ m = \frac{10 \times (2 \sqrt{3} + \sqrt{7}) }{5} \\ m = 2 \times (2 \sqrt{3} + \sqrt{7} ) \\ m = 4 \sqrt{3} + 2 \sqrt{7} [/tex]

Jawaban : A. 4√3 + 2√7

Maaf Kalau Salah

[tex] \frac{2 + \sqrt{8} }{ \sqrt{6} } \\ = \frac{2 + \sqrt{8} }{ \sqrt{6} } \times \frac{ \sqrt{6} }{ \sqrt{6} } \\ = \frac{2 \sqrt{6} + \sqrt{48} }{6} \\ = \frac{2 \sqrt{6} + 4 \sqrt{3} }{6} \\ = \frac{1}{3} \times ( \frac{ 2\sqrt{6} + 2 \sqrt{3} }{2} )[/tex]
[tex]m \times (2 \sqrt{3} - \sqrt{7} = 10 \\ m = \frac{10}{2 \sqrt{3} - \sqrt{7} } \\ m = \frac{10 \times (2 \ \sqrt{3} - \sqrt{7}) }{(2 \sqrt{3 } - \sqrt{7} ) {}^{2} } \\ m = \frac{10 \times (2 \sqrt{3} - \sqrt{7}) }{4(3) - 7} \\ m = \frac{10 \times (2 \sqrt{3} - \sqrt{7} }{5} \\ m = 2 (2 \sqrt{3} - \sqrt{7}) \\ m = 4 \sqrt{ \sqrt{3} } - 2 \sqrt{7} [/tex]