SOAL OSN DAN PEMBAHASAN
BAGIAN 2
1. Jika
x2 = yx – 1 dan y2 = 1 – y, maka x4
+ x3 + x2 + x
+ 1 = ….
Jawab :
x4 = (yx – 1)2 → x4
= y2x2 – 2yx + 1
y2x2 – 2yx
+ 1 + x(yx – 1) + yx – 1 + x + 1 → y2x2 – yx + yx2
+ 1
(1-y)x2 – yx + yx2
+ 1 →
x2 – yx + 1 = x2 – (yx-1) = 0
2. Jika
persamaan x3 – x + 1 = 0 memiliki akar-akar a, b, c, maka nilai
dari a8 + b8
+ c8 = .....
Jawab :
a+b+c = 0, a.b.c = -1, ab+bc+ac = -1
a2 + b2 + c2 = (a+b+c)2
– 2(ab+bc+ac)
a3 + b3 + c3 = 3abc + 3/2(a+b+c)( a2 + b2 + c2) - 1/2(a+b+c)3
a4 + b4 + c4 = (a2 + b2
+ c2)2 – 2{(ab)2+(bc)2+(ac)2}
{(ab)2+(bc)2+(ac)2} = (ab+bc+ac)2
– 2abc(a+b+c)
(ab)4 + (bc)4 + (ac)4 = {(ab)2+(bc)2+(ac)2}2
– 2(abc)2(a2 + b2 + c2)
a8 + b8 + c8 = (a4 + b4
+ c4)2 – 2{(ab)4 + (bc)4 + (ac)4}=
22 – 2(-3) = 10
3. Diketahui
P(x) = x3 – 3x2 + 1mempunyai akar-akarnya x1,
x2, x3. Jikaq(x) = x2 + 1, maka
tentukanlah nilai dari q(x1).q(x2).q(x3).
Jawab :
q(x1).q(x2).q(x3)
= (x12+1) (x22+1) (x32+1)
= {(x1.x2)2 + (x12+x22)+1}(x32+1)
(x1. x2. x3)2 + x32(x12+x22)
+ x32 + (x1.x2)2 + (x12+x22)
+ 1
1 + (x1+x2+x3)2 – 2(x1.x2+x2.x3+x1.x3)
+ (x1.x2+x2.x3+x1.x3)2
– 2 x1.x2.x3(x1 + x2 + x3)
+ 1
2 + 9 – 2.0 + 0
– 2.-1.3 = 11 + 6 = 17
4. Jika P(x) = x4 + 2x3 +2x2 – 1 akar-akarnya a, b, c, d dan Q(x) = x6 + x5 + x4 + x2 + x + 1,
maka tentukanlah nilai dari Q(a) + Q(b) + Q(c) +Q(d)
4. Jika P(x) = x4 + 2x3 +2x2 – 1 akar-akarnya a, b, c, d dan Q(x) = x6 + x5 + x4 + x2 + x + 1,
maka tentukanlah nilai dari Q(a) + Q(b) + Q(c) +Q(d)
Jawab :
ax4 + bx3 + cx2 +dx + e akar-akarnya x1,
x2, x3, x4, maka
x1 + x2 + x3 + x4 = - b/a, x1.
x2. x3. x4 = e/a
x1.x2.x3 + x1.x2.x4
+ x1.x3.x4 + x2.x3.x4
= - d/a
x1.x2 + x1.x3 + x1.x4
+ x2.x3 + x2.x4 + x3.x4
= c/a
Q(x) = (x2 – x + 1).P(x) + 2
Q(a)
+ Q(b) + Q(c) +Q(d) = 2 + 2 + 2 + 2 = 8 
Tidak ada komentar:
Posting Komentar