Jawaban:
Mencari nilai x:
[tex] \displaystyle \rm \large{x}^{2} + \frac{1}{ {x}^{2} } = 3[/tex]
[tex] \displaystyle \rm \large{x}^{2} + \frac{1}{ {x}^{2} } = 5 - 2[/tex]
[tex] \displaystyle \rm \large{x}^{2} + 2 + \frac{1}{ {x}^{2} } = 5[/tex]
[tex] \displaystyle \rm \large({x + \frac{1}{ x }})^{2} = 5[/tex]
[tex] \displaystyle \rm \large x + \frac{1}{x} = \sqrt{5} [/tex]
[tex] \displaystyle \rm \large x = \frac{ \sqrt{5} - 1 }{2},\:\frac{ \sqrt{5} + 1}{2} [/tex]
Uji coba nilai x:
untuk [tex] x = \frac{ \sqrt{5} - 1 }{2}[/tex]
[tex]\displaystyle \large \rm {(x)}^{3} = 2 + \sqrt{5} [/tex]
[tex]\displaystyle \large \rm {( \frac{ \sqrt{5} - 1 }{2} )}^{3} = 2 + \sqrt{5} [/tex]
[tex]\displaystyle \large \rm \sqrt{5 } - 2 = 2 + \sqrt{5}~~(tidak~memenuhi)[/tex]
untuk [tex]x = \frac{ \sqrt{5} + 1 }{2}[/tex]
[tex]\displaystyle \large \rm {( \frac{ \sqrt{5} + 1 }{2} )}^{3} = 2 + \sqrt{5} [/tex]
[tex]\displaystyle \large \rm2 + \sqrt{5} = 2 + \sqrt{5}~~(\bold{memenuhi})[/tex]
Sehingga:
[tex]\displaystyle \large \rm x = \frac{ \sqrt{5} + 1 }{2}[/tex]
[tex] \displaystyle \rm \large\frac{1}{x} = \frac{2}{ \sqrt{5} + 1} [/tex]
Maka:
[tex] \displaystyle \rm \large {x}^{2} + \frac{1}{ {x}^{2} } = 3[/tex]
[tex] \displaystyle \rm \large {x}^{2} + {(\frac{1}{ x})}^{2} = 3[/tex]
[tex] \displaystyle \rm \large{ (\frac{ \sqrt{5} + 1 }{2}) }^{2} + {(\frac{2}{ \sqrt{5} + 1 })}^{2} = 3[/tex]
[tex] \displaystyle \rm \large\frac{3 + \sqrt{5} }{2} + \frac{4 }{6+2\sqrt{5}} = 3[/tex]
[tex] \displaystyle \rm \large\frac{3 + \sqrt{5} }{2} + \frac{3 - \sqrt{5} }{2} = 3[/tex]
[tex] \displaystyle \rm \large \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2} = 3[/tex]
[tex] \displaystyle \rm \large\frac{6}{2} = 3[/tex]
[tex] \displaystyle \rm \large3 = 3[/tex]
Terbukti!
Penjelasan dengan langkah-langkah:
x² + 1/x² = 3
x² + 2 + 1/x³ = 5
x + 1/x = √5
x = √5 - 1/2
maka
x² +( 1/x)² = 3
3 - √5/2 + 4/6 - 2√5 = 3
6/2 = 3
3 = 3
terbukti
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Jawaban:
Mencari nilai x:
[tex] \displaystyle \rm \large{x}^{2} + \frac{1}{ {x}^{2} } = 3[/tex]
[tex] \displaystyle \rm \large{x}^{2} + \frac{1}{ {x}^{2} } = 5 - 2[/tex]
[tex] \displaystyle \rm \large{x}^{2} + 2 + \frac{1}{ {x}^{2} } = 5[/tex]
[tex] \displaystyle \rm \large({x + \frac{1}{ x }})^{2} = 5[/tex]
[tex] \displaystyle \rm \large x + \frac{1}{x} = \sqrt{5} [/tex]
[tex] \displaystyle \rm \large x = \frac{ \sqrt{5} - 1 }{2},\:\frac{ \sqrt{5} + 1}{2} [/tex]
Uji coba nilai x:
untuk [tex] x = \frac{ \sqrt{5} - 1 }{2}[/tex]
[tex]\displaystyle \large \rm {(x)}^{3} = 2 + \sqrt{5} [/tex]
[tex]\displaystyle \large \rm {( \frac{ \sqrt{5} - 1 }{2} )}^{3} = 2 + \sqrt{5} [/tex]
[tex]\displaystyle \large \rm \sqrt{5 } - 2 = 2 + \sqrt{5}~~(tidak~memenuhi)[/tex]
untuk [tex]x = \frac{ \sqrt{5} + 1 }{2}[/tex]
[tex]\displaystyle \large \rm {( \frac{ \sqrt{5} + 1 }{2} )}^{3} = 2 + \sqrt{5} [/tex]
[tex]\displaystyle \large \rm2 + \sqrt{5} = 2 + \sqrt{5}~~(\bold{memenuhi})[/tex]
Sehingga:
[tex]\displaystyle \large \rm x = \frac{ \sqrt{5} + 1 }{2}[/tex]
[tex] \displaystyle \rm \large\frac{1}{x} = \frac{2}{ \sqrt{5} + 1} [/tex]
Maka:
[tex] \displaystyle \rm \large {x}^{2} + \frac{1}{ {x}^{2} } = 3[/tex]
[tex] \displaystyle \rm \large {x}^{2} + {(\frac{1}{ x})}^{2} = 3[/tex]
[tex] \displaystyle \rm \large{ (\frac{ \sqrt{5} + 1 }{2}) }^{2} + {(\frac{2}{ \sqrt{5} + 1 })}^{2} = 3[/tex]
[tex] \displaystyle \rm \large\frac{3 + \sqrt{5} }{2} + \frac{4 }{6+2\sqrt{5}} = 3[/tex]
[tex] \displaystyle \rm \large\frac{3 + \sqrt{5} }{2} + \frac{3 - \sqrt{5} }{2} = 3[/tex]
[tex] \displaystyle \rm \large \frac{3 + \sqrt{5} + 3 - \sqrt{5} }{2} = 3[/tex]
[tex] \displaystyle \rm \large\frac{6}{2} = 3[/tex]
[tex] \displaystyle \rm \large3 = 3[/tex]
Terbukti!
Penjelasan dengan langkah-langkah:
x² + 1/x² = 3
x² + 2 + 1/x³ = 5
x + 1/x = √5
x = √5 - 1/2
maka
x² +( 1/x)² = 3
3 - √5/2 + 4/6 - 2√5 = 3
6/2 = 3
3 = 3
terbukti