Jawab:a. 5
Penjelasan :
[tex]\displaystyle\frac{\frac{1}{2}-\frac{1}{\sqrt5}}{\frac{1}{2}+\frac{1}{\sqrt5}}\\\\=\frac{\frac{\sqrt5}{2\sqrt5}-\frac{2}{2\sqrt5}}{\frac{\sqrt5}{2\sqrt5}+\frac{2}{2\sqrt5}}\\\\=\frac{\frac{\sqrt5-2}{2\sqrt5}}{\frac{\sqrt5+2}{2\sqrt5}}\\\\=\frac{\sqrt5-2}{\sqrt5+2}\\\\=\frac{\sqrt5-2}{\frac{1}{\sqrt5-2}}\\\\=(\sqrt5-2)^2\\=(\sqrt{5})^2-2(2)\sqrt{5}+2^2\\=5-4\sqrt{5}+4\\=9-4\sqrt{5}\\\rightarrow a+b\sqrt{5}\\a=9,\:\:b=-4\\a+b=9+(-4)\\~~~~~~~\:= 9-4\\~~~~~~~\:= 5[/tex]
(xcvi)
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Verified answer
Jawab:
a. 5
Penjelasan :
[tex]\displaystyle\frac{\frac{1}{2}-\frac{1}{\sqrt5}}{\frac{1}{2}+\frac{1}{\sqrt5}}\\\\=\frac{\frac{\sqrt5}{2\sqrt5}-\frac{2}{2\sqrt5}}{\frac{\sqrt5}{2\sqrt5}+\frac{2}{2\sqrt5}}\\\\=\frac{\frac{\sqrt5-2}{2\sqrt5}}{\frac{\sqrt5+2}{2\sqrt5}}\\\\=\frac{\sqrt5-2}{\sqrt5+2}\\\\=\frac{\sqrt5-2}{\frac{1}{\sqrt5-2}}\\\\=(\sqrt5-2)^2\\=(\sqrt{5})^2-2(2)\sqrt{5}+2^2\\=5-4\sqrt{5}+4\\=9-4\sqrt{5}\\\rightarrow a+b\sqrt{5}\\a=9,\:\:b=-4\\a+b=9+(-4)\\~~~~~~~\:= 9-4\\~~~~~~~\:= 5[/tex]
(xcvi)