1. Sederhanakan penyebut dari bentuk akar berikut; A. 5 per √2 B. 6 per 2√3 C. -4 per √10 D. √2 per √11 E. -3√6 per √5 F. 7 per 3√2 G. 4√9 per √8 H. √3 per 3√25

Jawaban dapat dilihat pada pembahasan.

Pembahasan

A. 5 per √2

[tex]\frac{5}{\sqrt{2}}=\frac{5}{\sqrt{2} }\times\frac{\sqrt{2}}{\sqrt{2}}\\\\\frac{5}{\sqrt{2}}=\frac{5\sqrt{2} }{\sqrt{2}\times\sqrt{2}}\\\\\frac{5}{\sqrt{2}}=\frac{5\sqrt{2}}{2}\\\\\frac{5}{\sqrt{2}}=\frac{5}{2}\sqrt{2}[/tex]

B. 6 per 2√3

[tex]\frac{6}{2\sqrt{3}}=\frac{6}{2\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}\\\\\frac{6}{2\sqrt{3}}=\frac{6\sqrt{3} }{2\sqrt{3}\times\sqrt{3}}\\\\\frac{6}{2\sqrt{3}}=\frac{6\sqrt{3} }{2\times{3}}\\\\\frac{6}{2\sqrt{3}}=\frac{6\sqrt{3} }{6}\\\\\frac{6}{2\sqrt{3}}=\sqrt{3}[/tex]

C. -4 per √10

[tex]-\frac{4}{\sqrt{10}}=-\frac{4}{\sqrt{10}}\times\frac{\sqrt{10}}{\sqrt{10}}\\\\-\frac{4}{\sqrt{10}}=-\frac{4\sqrt{10}}{\sqrt{10}\times\sqrt{10}}\\\\-\frac{4}{\sqrt{10}}=-\frac{4\sqrt{10}}{10}\\\\-\frac{4}{\sqrt{10}}=-\frac{2}{5}\sqrt{10}[/tex]

D. √2 per √11

[tex]\frac{\sqrt{2}}{\sqrt{11}}= \frac{\sqrt{2}}{\sqrt{11}}\times\frac{\sqrt{11}}{\sqrt{11}}\\\\\frac{\sqrt{2}}{\sqrt{11}}= \frac{2\sqrt{11}}{\sqrt{11}\times\sqrt{11}}\\\\\frac{\sqrt{2}}{\sqrt{11}}= \frac{2\sqrt{11}}{11}\\\\\frac{\sqrt{2}}{\sqrt{11}}= \frac{2}{11}\sqrt{11}[/tex]

E. -3√6 per √5

[tex]-\frac{3\sqrt{6}}{\sqrt{5}}=-\frac{3\sqrt{6}}{\sqrt{5}}\times\frac{\sqrt{5}}{\sqrt{5}}\\ \\-\frac{3\sqrt{6}}{\sqrt{5}}=-\frac{3\sqrt{6}\times\sqrt{5}}{\sqrt{5}\times\sqrt{5}}\\\\-\frac{3\sqrt{6}}{\sqrt{5}}=-\frac{3\sqrt{30}}{5}\\\\-\frac{3\sqrt{6}}{\sqrt{5}}=-\frac{3}{5}\sqrt{30}[/tex]

F. 7 per 3√2

[tex]\frac{7}{3\sqrt{2}}=\frac{7}{3\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\\\\frac{7}{3\sqrt{2}}=\frac{7\sqrt{2}}{3\sqrt{2}\times\sqrt{2}}\\\\\frac{7}{3\sqrt{2}}=\frac{7\sqrt{2}}{3\times{2}}\\\\\frac{7}{3\sqrt{2}}=\frac{7\sqrt{2}}{6}\\\\\frac{7}{3\sqrt{2}}=\frac{7}{6}\sqrt{2}[/tex]

G. 4√9 per √8

[tex]\frac{4\sqrt{9}}{\sqrt{8}}=\frac{4\times{3}}{\sqrt{4\times{2}}}\\\\\frac{4\sqrt{9}}{\sqrt{8}}=\frac{12}{2\sqrt{2}}\\\\\frac{4\sqrt{9}}{\sqrt{8}}=\frac{12}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\ \\\frac{4\sqrt{9}}{\sqrt{8}}=\frac{12}{2\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}\\ \\\frac{4\sqrt{9}}{\sqrt{8}}=\frac{12\sqrt{2} }{2\times{2}}\\\\\frac{4\sqrt{9}}{\sqrt{8}}=\frac{12\sqrt{2}}{4}\\\\{\sqrt{8}}=3\sqrt{2}[/tex]

H. √3 per 3√25

[tex]\frac{\sqrt{3} }{3\sqrt{25}}= \frac{\sqrt{3} }{3\times{5}}\\\\\frac{\sqrt{3} }{3\sqrt{25}}= \frac{\sqrt{3} }{15}\\\\\frac{\sqrt{3} }{3\sqrt{25}}= \frac{1}{15}\sqrt{3}[/tex]

2. a

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

[tex]\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{2\sqrt{6}((\sqrt{2}+\sqrt{3})-\sqrt{5})}{2\sqrt{6}}\\\\\frac{2\sqrt{6}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\sqrt{2}+\sqrt{3}-\sqrt{5}[/tex]

b.

[tex]\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\frac{\sqrt{11}-2\sqrt{30}+1}{\sqrt{6}-\sqrt{5}-2\sqrt{6}}\\\\\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\frac{\sqrt{11}-2\sqrt{30}+1}{-\sqrt{5}-\sqrt{6}}\times{\frac{-\sqrt{5}+\sqrt{6}}{-\sqrt{5}+\sqrt{6}}}\\\\\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\frac{\sqrt{11}-2\sqrt{30}+1(-\sqrt{5}+\sqrt{6})}{(-\sqrt{5})^2-(\sqrt{6})^2}[/tex]

[tex]\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\frac{\sqrt{11}-2\sqrt{30}+1(-\sqrt{5}+\sqrt{6})}{5-6}\\\\\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\frac{\sqrt{11}-2\sqrt{30}+1(-\sqrt{5}+\sqrt{6})}{-1}\\\\\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=(-1)(-\sqrt{55}+\sqrt{66}+2\sqrt{150}-2\sqrt{180}-\sqrt{5}+\sqrt{6})\\\\\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\sqrt{55}-\sqrt{66}-2.5\sqrt{6}+2.6\sqrt{5}+\sqrt{5}-\sqrt{6}\\[/tex]

[tex]\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\sqrt{55}-\sqrt{66}-10\sqrt{6}+12\sqrt{5}+\sqrt{5}-\sqrt{6}\\\\\frac{\sqrt{11}-\sqrt{120}+1}{\sqrt{6}-\sqrt{5}-\sqrt{24}}=\sqrt{55}-\sqrt{66}-11\sqrt{6}+13\sqrt{5}[/tex]

c. tidak bisa diselesaikan karena soal tidak lengkap.

Pelajari tentang merasionalkan bentuk akar pada :

brainly.co.id/tugas/11959478

brainly.co.id/tugas/23275379

brainly.co.id/tugas/23394227

Detail Jawaban

Kelas : 10

Mapel : Matematika

Materi : Bentuk Akar, Eksponen, Logaritma

Kode kategorisasi : 10.2.1.1

Kata kunci : bentuk akar, pangkat, rasional, pecahan.