Jika ab = 1, bc = 2, cd = 3, de = 4 dan ea = 6 serta a, b, c, d dan e bilangan positif, maka a + b + c + d + e adalah
Matematika
farhan59
Pertanyaan
Jika ab = 1, bc = 2, cd = 3, de = 4 dan ea = 6 serta a, b, c, d dan e bilangan positif, maka a + b + c + d + e adalah
2 Jawaban
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1. Jawaban ouja
[tex]ab=1[/tex] ⇒ [tex]b= \frac{1}{a} [/tex]
[tex]bc=2[/tex] ⇒ [tex]c= \frac{2}{b}= \frac{2}{ \frac{1}{a} }=2a[/tex]
[tex]cd=3[/tex] ⇒ [tex](2a)d=3=\frac{3}{2a}[/tex]
[tex]de=4[/tex] ⇒ [tex] \frac{3}{2}a(e)=4[/tex]
[tex]e= \frac{4}{ \frac{3}{2}a } = \frac{8}{3}a [/tex]
[tex]ea=6[/tex] ⇒ [tex]( \frac{8}{3}a )a=6[/tex]
[tex] a^{2} = \frac{6}{ \frac{8}{3} }=\frac{9}{4}[/tex]
[tex]a= \frac{3}{2} [/tex]
[tex]b= \frac{1}{a}= \frac{1}{ \frac{3}{2} }= \frac{2}{3} [/tex]
[tex]c=2a=2 \frac{3}{2} =3[/tex]
[tex]d= \frac{3}{2}a= \frac{3}{2}. \frac{3}{2}= \frac{9}{4} [/tex]
[tex]e= \frac{8}{3}a= \frac{8}{3}. \frac{3}{2}=4[/tex]
[tex]a+b+c+d+e= \frac{3}{2}+ \frac{2}{3}+3+\frac{9}{4}+4= \frac{18+8+36+27+48}{12}=\frac{137}{12}[/tex] -
2. Jawaban rezamahesaa
ab = 1
b = 1/a
bc = 2
(1/a) c = 2
c/a = 2
c = 2a
cd = 3
(2a) d = 3
2ad = 3
d = 3/2a
de = 4
(3/2a) e = 4
e = 4 x 2a/3
e = 8a/3
ea = 6
(8a/3) a = 6
8a^2 / 3 = 6
8a^2 = 6 x 3
8a^2 = 18
a^2 = 18 / 8
a^2 = 9 / 4
a = 3/2
a sudah dapat, tinggal cari yang lain
ab = 1
(3/2) b = 1
b = 2/3
bc = 2
(2/3) c = 2
c = 2 x 3/2
c = 3
cd = 3
(3) d = 3
d = 3 / 3
d = 1
de = 4
(1) e = 4
e = 4
a + b + c + d + e
= 3/2 + 2/3 + 3 + 1 + 4
= 9/6 + 4/6 + 8
= 13/6 + 48/6
= 61/6