1. 2+4+6+...+2n = n(n+1)1 n = 1;
2 = 1(1+1)
2 = 2
2 n = k;
2+4+6+...+2k = k(k+1)
3 n = k+1;
2+4+6+...+2(k+1) = (k+1)(k+1+1)
2(k+1) = (k+1)(k+2)
2+4+6+...+2k+2(k+1) = k(k+1)+2(k+1)
= (k+1)(k+2)
2. 13+24+35+...+n(n+2) = 1 n = 1;
1(1+2) = 13 = 3 = 32 n = k;
13+24+35+...+k(k+2) = 3 n = k+1;
13+24+35+...+(k+1)(k+1+2) = (k+1)(k+3) =
13+24+35+...+k(k+2)+(k+1)(k+3) = + (k+1)(k+3) =
=
= = = = = 3. 57 +1 n = 1;
57 + 57 + 57 343+512
57 855
15
2 n = k;
57 +
A
3 n = k+1;
57 + 57 + 57 7+64 57 7+(57+7) B
57 7+57+7 57 7(+)+57B
57 7A+57B4. 2x+1-3-x = 1-2x
2x+1 = 0 3-x = 0 2x = -1 x = 3
x = - 1 ( -, - ) -(2x+1)-(3-x) = 1-2x
-2x-1-3+x = 1-2x
-x-4 = 1-2x
x = 5 nije rjeenje
2 ( - , 3 )
2x+1-(-(3-x)) = 1-2x
2x+1+3-x = 1-2x
x+4 = 1-2x
5x = 3
x = rjeenje3 (3, + )
2x+1-(-(3-x)) = 1-2x
2x+1+3-x = 1-2x
x+4 = 1-2x
3x = 3
x = 1 nije rjeenje
5. x-1-2x+1 = x-3
x-1 = 0 2x+1 = 0
x = 1 2x = -1
x = - 1 ( -, - )
-(x-1)-(-(2x+1) = x-3
-x+1+2x+1 = x-3
x+2 = x-3 nema rjeenja
2 ( -, - )
-(x-1)-(2x+1) = x-3
-x+1-2x-1 = x-3
-3x-x = -3
-4x = -3
x = rjeenje
3 (1, + ) (x-1)-(2x+1) = x-3
x-1-2x-1 = x-3
-x-2 = x-3
-2x = -1
x = nije rjeenje6. Odrediti imaginarni i realni dio, te nai z iz
z = .z = = = = = n= = = = Realni dio: a = 0
Imaginarni dio: b = z = = = = = 7. Odrediti apsulutno z . u =
1-2i
v = 2-i
(1-2i) = 1-31(2i)+3(2i)-(2i) = 1-6i+12 i-8 i = 1-6i-12+8i =
-11+2i
(2-i) = 2-32i+32i-i = 8-12i-6+i = 2-11iz = = = = = = n= = = = -
- iRealni dio: a = - Imaginarni dio: b = - z = = = = = 8. Razloiti
na elementarne razlomke slijedee racionalne funkcije!f(x) = = 0
x1,2 = x1,2 = = =
x1 = 2 x2 = - = + (3x+1)(x-2)
11x-1 = A(x-2)+B(3x+1)11x-1 = Ax-2A+3Bx+B
11x-1 = (A+3B)x-2A+BA+3B = 11 A = 11-3B A = 11-33 = 11-9 = 2
-2A+B = -1
-2(11-3B)+B = -1
-22+6B+B = -1
7B = 21
B = 3 = + = + 9. f(x) = = + + (x-1)(x+1)x-2 =
A(x-1)(x+1)+B(x+1)+(Cx+D)(x-1)
x-2 = Ax+Ax-Ax-A+Bx+B+Cx-2Cx+Cx+Dx-2Dx-2D+D
x-2 = x(A+C)+x(-A+B-2C+D)+x(A+C-2D)+(-A+B+D)
(1) A+C = 0 A+C-2D = 0 -A+B+D = -2
(2) -A+B-2C+D = 1 -(A+C) = 0__ -(-A+B-2C+D) = 1(3) A+C-2D = 0
-2D = 0 2C = -3(4) -A+B+D = -2 D = 0 C = - (1)A+C = 0 (4) -A+B+D =
-2 A - = 0 - +B = -2 2
A = -3+2B = -4
2B = -1
B = - = + + = - - 10. Broj x =i je jedno rjeenje jednaine
x-2x+9x-6x+6. Odrediti ostala rjeenja!x1 = ix2 = - i(x- x1)(x- x2)
= (x- i)(x+i) = x-(i) = x+3
x-2x+9x-6x+6:(x+3) = x-2x+2
-x-3x -2x+6x-6x+6
- 2x 6x x-2x+2 = 0 +6x+6 x3,4 = +6x+6 x3,4 = x3,4 = = x3,4 =
1i
x3 = 1+i x4 = 1-i
11. Ako je x1 = -1-i jedno rjeenje jednaine x-5x-10x-6, odrediti
ostala rjeenja a zatim izraunati x1+x2+x3+x4.x1 = -1+i
x2 = -1-i
(x- x1)(x- x2) = (x-(-1+i))(x-(-1-i)) =(x+1-i)(x+1+i) = (x+1)-i
= = x+2x+1+1 = x+2x+2
(x-5x-10x-6):( x+2x+2) = x-2x-3
- x2x2x
-2x-7x-10x-6
-(-2x-4x-4x)
-3x-6x-6
-(-3x-6x-6)
x-2x-3 = 0
x3,4 = x3,4 = = x3 = = 3 x4 = = -1
x1+x2+x3+x4 = (-1+i)+(-1-i)+3+(-1) = 2+2i+2-2i+27+3 = 34
(-1+i) = (-1) + 3(-1)i+3(-1)i+i = -1+3i+3-i = 2+2i
(-1-i) = (-1) - 3(-1)i+3(-1)i-i = -1-3i+3+i = 2-2i
12. Ako su x1 = -2 i x2 = 4 nule polinoma x-2x-8x-8x+16x+64 = 0,
odrediti ostale nule tog polinoma!(x- x1)(x- x2) =(x+2)(x- 4) =
x-4x+2x-8 = x-2x-8(x-2x-8x-8x+16x+64):x = 2x-8 = x-8
-( x-2x-8x)
-8x+16x+64
-(-8x+16x+64)
x-8 = 0 x1 = -2
x = 8 x2 = 4
x3,4,5 = x3 = 2
x3,4,5 = 2 x4 = 2
x5 = 2
13. algebarski oblik: z = a+yi trigonometrijski oblik: z = rcis
= r(cos+i sin)
x y
z = tg = = = + = r = = = = = 1z = 1ciswk = cis
k = 0; 1; 2; 3
wk = w0 = 1 cis = = 1cis = 1(cos +sin i) = + w1 = 1 cis = 1 cis
= cis = cis = + w2 = 1 cis = 1 cis = cis = cis = - w3 = 1 cis = 1
cis = cis = cis = - 7
