Tolong soal olimpiade

Tolong soal olimpiade

Jawaban Terkonfirmasi

$largetext{$begin{aligned}sf Jawaban: boxed{:bf1:}end{aligned}$}$
 

Pembahasan

Diketahui
Untuk bilangan real positif a, b, c, p, q, dan r berlaku:
$begin{aligned}frac{a}{p}=frac{b}{q}=frac{c}{r}end{aligned}$

Ditanyakan
$begin{aligned}frac{abccdot(p+q)(q+r)(r+p)}{pqrcdot(a+b)(b+c)(c+a)}={dots}end{aligned}$

PENYELESAIAN

$begin{aligned}&frac{a}{p}=frac{b}{q}=frac{c}{r}\&{implies}begin{cases}(1)&aq=bp\(2)&br=cq\(3)&ar=cpend{cases}end{aligned}$

Kemudian, kita selesaikan.

$begin{aligned}&frac{abccdot(p+q)(q+r)(r+p)}{pqrcdot(a+b)(b+c)(c+a)}\{= }&frac{a(p+q)cdot b(q+r)cdot c(r+p)}{p(a+b)cdot q(b+c)cdot r(c+a)}\{= }&frac{(ap+aq)}{(ap+bp)}cdotfrac{(bq+br)}{(bq+cq)}cdotfrac{(cr+cp)}{(cr+ar)}end{aligned}$
$begin{aligned}{= }&frac{overbrace{(ap+bp)}^{!!!begin{array}{c}(1):bpto aqend{array}!!!}}{(ap+bp)} cdot frac{overbrace{(bq+cq)}^{!!!begin{array}{c}(2):cqto brend{array}!!!}}{(bq+cq)} cdot frac{overbrace{(cr+ar)}^{!!!begin{array}{c}(3):arto cpend{array}!!!}}{(cr+ar)}\{= }&1cdot1cdot1\{= }&boxed{:bf1:}end{aligned}$

KESIMPULAN

$begin{aligned}therefore frac{abccdot(p+q)(q+r)(r+p)}{pqrcdot(a+b)(b+c)(c+a)}=boxed{:bf1:}end{aligned}$
$blacksquare$
 