數學知識交流 － 數學雜題 (2)

5.
[sin(3π-θ)]/[tan(π/2+θ)] [cos(4π-θ)]/[sin(-θ-2π)]
= [sin((3π-θ)-2π)]/[-1/tanθ] [cos(-θ)]/[sin(2π+(-θ-2π))]
= [sin(π+θ)]/[-cosθ/sinθ] [cosθ]/[sin(-θ)]
= [-cosθ] [-sinθ/cosθ] [cosθ]/[-sinθ]
= [-cosθ]

6a.
x = 3-7 = -4 => y = 3 - (-4) = 7
6b.
y - 8 = 9 => y = 9+8 = 17
6c.
x = y + 4
y^2 - (y+4) = 9
y^2 - y - 4 = 9
y^2 - y - 13 = 0 => y = {-(-1) + sqrt[1+4(1)(13)]} / 2(1)
or y = {-(-1) - sqrt[1+4(1)(13)]} / 2(1)
=> y = {1+sqrt[53]} / 2 or {1-sqrt[53]} / 2
6d.
(a + b + c + d) + y = 7 and a + b + c + d = 7
7 + y = 7 => y = 0
6e.
y = -2-(-9+(-8))
y = -2-(-9-8)
y = -2-(-17)
y = -2+17 => y = 15
6f.
x = 9 and u = x^2 => u = 81
81 = y + 45 => y = 36
6g.
b + 3 = 5 => b = 2
a = b + 3 => a = 5
a = z = x = y = 5
6h.
y = [(1987!)^29]^0 = 1
as any real number to power "0" = 1
A^0 = A^(x-x) = A^x / A^x = 1

7.
ball = 59x
pen = 6y
$: 59x + 6y = 1009
items: x + y = 18
[(x+y) + 6] / 2 = [18+6] / 2 = 12

8a.
x^2 - 1 > 0
(x+1)(x-1) > 0 implies there will be two cases
case i,
x+1 > 0 and x-1 > 0 => x > -1 and x > 1 => x > 1
case ii,
x+1 < 0 and x-1 < 0 => x < -1 and x < 1 => x < -1
thus, x < -1 or x > 1

8b.
x^3 + 3x^2 + 3x + 1 > 0
(x + 1)^3 > 0 implies there will be only one case
x+1 > 0 => x > -1

8c.
(x-1)(x-2)(x-3)(x-4) < 0, separate into 2 major terms
case i,
(x-1)(x-2) > 0 and (x-3)(x-4) < 0
...
[x < 1 or 2 < x] and [3 < x < 4]
3 < x < 4
case ii,
(x-1)(x-2) < 0 and (x-3)(x-4) > 0
...
[1 < x < 2] and [x < 3 or 4 < x]
1 < x < 2
thus, 1 < x < 2 or 3 < x < 4

2011-11-30 12:34:52 補充：
5. [sin(3π-θ)] = [sin(π-θ)] = [sinθ]
...
Ans = cosθ

(5) (-cosθ/-cotθ) x ( cosθ/sinθ) = cosθ
(6) (a.) y= 7
(b) y = 17
(c) y = (1±√53) /2
(d) y = 0
(e) y = 15
(f) y = 36
(g) y = 5
(h) y = 1

(7) x = 17 and y = 1 ∴ [(x + y) + 6]/2 = 12

(8) (a) x < -1 , x > 1
(b) x > -1
(c) 1 < x < 2 , 3 < x < 4

6b, y is 18 that all