hm.
[tex]\bf~jika~ \sqrt{x + 15} = \sqrt{x} + 3, \\ \\ \bf~maka~ {\sqrt{ \sqrt{ \sqrt{25x}. \sqrt{x} } } }^{4} = ....[/tex]
.....
Jawab:
5
Penjelasan dengan
langkah-langkah:
[tex]\bf\displaystyle \sqrt{x+15}=\sqrt{x}+3\\x+15=(\sqrt{x}+3)^2\\x+15=x+6\sqrt{x}+9\\6\sqrt{x}=15-9\\6\sqrt{x}=6\\\sqrt{x}=6\div6\\\sqrt{x}=1\\x=1[/tex]
maka
[tex]\displaystyle\bf \sqrt{\sqrt{\sqrt{25x}\cdot\sqrt{x}}}^4=\\\sqrt{\sqrt{\sqrt{25(1)}\cdot\sqrt{1}}}^4=\\\sqrt{\sqrt{\sqrt{25}}}^4=\sqrt{\sqrt{\sqrt{5^2}}}^4\\=\sqrt{\sqrt{5}}^4=5^{(1/4)\cdot4}=[/tex]
5
( x c v i )
[answer.2.content]
