jika m dan n akar akar persamaan kuadarat 25x^2-6x-12=0 maka nilai dari: a.1÷m + 1÷n b.n÷m+m÷n c.m^2+n^2

25x^2 - 6x -12 = 0
a=25,b=-6,c=-12

m+n = -a/b = 6/25
mxn = c/a = -12/25

a. 1/m + 1/n = n/mn + m/mn = (m+n)/mn = (6/25)/(-12/25) = 6/25 x -25/12 = -6/12 = -1/2

b n/m + m/n
= (n*n/m*n) + (m*m/n*m)
= n^2/mn + m^2/mn
= (m^2 + n^2)/mn
= (m+n)^2 - 2mn/mn
= (6/25)^2 - 2(-12/25) /(-12/25)
= 36/625 +24/25/(-12/25)
= 36/625 + 600/625/(-12/25)
= 636/625 x 25/-12
= -636/300
= -53/12

c. m^2 + n^2
= (m+n)^2 - 2.mn
= (6/25)^2 -2(-12/25)
= 36/625 + 24/25
= 36/625 + 600/625
= 636/625

Mapel : Matematika
Kelas : X SMA
Bab : Persamaan Kuadrat

Pembahasan :
25x² - 6x - 12 = 0

Berdasarkan Teorema Vietta
m + n = -b/a
m + n = 6/25

m . n = c/a
m . n = -12/25

Maka :
1/m + 1/n
= (m + n)/mn
= (6/25)/(-12/25)
= - 1/2

n/m + m/n
= (n² + m²)/mn
= [(m + n)² - 2mn]/mn
= [(6/25)² - 2(-12/25)]/(-12/25)
= [36/625 + 24/25]/(-12/25)
= -53/25

m² + n²
= (m + n)² - 2mn
= (6/25)² - 2(-12/25)
= 36/625 + 24/25
= 36/625 + 600/625
= 636/625
= 1 (11/625)