Buktikan bahwa: [tex]\displaystyle \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{5}}=\frac{3}{2}\sqrt{2}+\frac{1}{2}\sqrt{30}-\frac{1}{2}\sqrt{15}-\frac{3}{2}[/tex]
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Buktikan bahwa:
[tex]\displaystyle \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{5}}=\frac{3}{2}\sqrt{2}+\frac{1}{2}\sqrt{30}-\frac{1}{2}\sqrt{15}-\frac{3}{2}[/tex]
[tex]\displaystyle \frac{\sqrt{3}-\sqrt{6}}{\sqrt{3}-\sqrt{5}}=\frac{3}{2}\sqrt{2}+\frac{1}{2}\sqrt{30}-\frac{1}{2}\sqrt{15}-\frac{3}{2}[/tex]
2 Jawaban
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1. Jawaban ShanedizzySukardi
Materi Bentuk Akar
<<<<<2. Jawaban wilsonrubik123
√3 - √6 / (√3 - √5) = (3/2)√2 + (1/2)√30 - (1/2)√15 - 3/2
(√6 - √3)(√5 + √3) / (√5 - √3)(√5 + √3) = (3√2 + √30 - √15 - 3) / 2
(√30 + √18 - √15 - √3^2) / (√5^2 - √3^2) = (3√2 + √30 - √15 - 3) / 2
(√30 + √18 - √15 - 3) / (5 - 3) = (3√2 + √30 - √15 - 3) / 2
(√30 + √18 - √15 - 3) / 2 = (3√2 + √30 - √15 - 3) / 2
1 = 1
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