Soal Matematika IPA SMA, a > b > 0 dan a^2 + b^2 = 8ab maka nilai a+b/a-b = .... Question from @Salsafiapf

hendrisyafa $a^{2}+b^{2}= 8ab$

$(a+b)^{2}-2ab=8ab$

$(a+b)^{2}=10ab$

$a+b= \sqrt{10ab}$

jika : $(a-b)^{2}=a^{2}+b^{2}-2ab$

maka :

$a-b= \sqrt{a^{2}+b^{2}-2ab}$

$a-b= \sqrt{8ab-2ab}$

$a-b= \sqrt{6ab}$ x (-1)

$b-a = - \sqrt{6ab}$

sehingga :

$\frac{a+b}{b-a}= \frac{ \sqrt{10ab} }{- \sqrt{6ab} }$

$\frac{a+b}{b-a}=- \sqrt{ \frac{10ab}{6ab} }$

$\frac{a+b}{b-a}=- \sqrt{ \frac{5}{3} }$

$\frac{a+b}{b-a}=- \frac{ \sqrt{5} }{ \sqrt{3} }$

$\frac{a+b}{b-a}=- \frac{ \sqrt{5} }{ \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} }$

$\frac{a+b}{b-a}=- \frac{ 1 }{ 3 } \sqrt{15}$

QUICK METHOD / CARA CEPAT

$\frac{a+b}{b-a}= \frac{a+b}{-(a-b)}$

$= -\sqrt{\frac{(a+b)^{2}}{(a-b)^{2}} }$

$= -\sqrt{\frac{(a^{2}+b^{2})+2ab}{(a^{2}+b^{2})-2ab} }$

$= -\sqrt{\frac{8ab+2ab}{8ab-2ab} }$

$= -\sqrt{\frac{10ab}{6ab} }$

$= -\sqrt{\frac{5}{3} }$

$=- \frac{1}{3} \sqrt{15}$

1 votes Thanks 1

Probie101 A^2+b^2=8ab
...
(a+b)^2-2ab=8ab
(a+b)^2=10ab
a+b=(10ab)^(1/2)
...
(b-a)^2=a^2+b^2-2ab
(b-a)^2=(8ab)-2ab
(b-a)^2=6ab
b-a=(6ab)^(1/2)
...
(a+b)/(b-a)=(10ab)^(1/2):(6ab)^(1/2)
(a+b)/(b-a)=10^(1/2):6^(1/2)
(a+b)/(b-a)=(10/6)^(1/2) atau
(5/3)^(1/2)=(1/3).(15)^(1/2)... (E)

0 votes Thanks 0

hendrisyafa Perhatikan syarat a>b>0. Artinya b-a akan menghasilkan negatif. Shg a²+b² /b-a akan menghasilkan negatif

Probie101 Oia, siap bro, kalem aja bro.. Biar belajar juga, orang ga selalu benar toh.. Thanks bro

Soal no. 58 tentang pertumbuhan dan perkembangan tumbuhan

Soal no. 58 tentang pertumbuhan dan perkembangan tumbuhan

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