(20分) Angles & circles + Algebraic fractions and equations

答了部分題目, 如果你不介意, 其實我稍後才答.
4.
Join BE,
In ΔABE,
∠EAB = ∠EBA = 45º (base ∠s, isos. Δ)
∠AEB = 180º - ∠EAB - ∠EBA (∠ sum of Δ)
= 180º - 45º - 45º
= 90º
∠BEC = 360º - ∠BEA - ∠AEC (∠s at a pt.)
= 360º - 90º - 160º
= 110º
In ΔBEC,
∠BCE = ∠EBC (base ∠s, isos. Δ)
∠BCE + ∠EBC + ∠BEC = 180º (∠ sum of Δ)
2∠BCE = 180º - 110º
∴∠BCE = 35º

∠ABC = ∠EBA + ∠EBC
= 45º + 35º
= 80º
∠ABC + ∠ADC = 180º (opp. ∠s, cyclic quad.)
∠ADC = 180º - 80º
∴∠ADC = 100º

9.
(a)
3/(x^2-9) + x/(x+3)
= 3/[(x-3)(x+3)] + x/(x+3)
= [3 + x(x-3)] / [(x-3)(x+3)]
= (3 + x^2 – 3x) / [(x-3)(x+3)]
= (x^2 – 3x + 3) / [(x-3)(x+3)]
= (x^2 – 3x + 3) / (x^2-9)

(b)
3x/(x+1) + x/(x-1) = 4
[(3x)(x-1)+x(x+1)] / [(x+1)(x-1)] = 4
(3x^2-3x+x^2+x) / [(x+1)(x-1)] = 4
(4x^2-2x) / [(x+1)(x-1)] = 4
[2x(2x-1)] / [(x+1)(x-1)] = 4
If you want to solve for x, then the following steps are needed:
2x(2x-1) = 4(x+1)(x-1)
4x^2-2x = 4x^2-4
-2x = -4
x = 2

(c)
1/(x-1) 1- (1/x) = 8
[(x)-(x-1)] / [x(x-1)] = 8
1 / [x(x-1)] = 8
If you want to solve for x, then the following steps are needed:
1 = 8x(x-1)
1 = 8x^2-8
8x^2-7 = 0
x = ±√(7/8)

16.
Mistake 1:
In Line 2,
After removing the brackets, -4 x -2 = 8
So, the correct step is (6x + 12 – 4x + 8)/[(x – 2)(x + 2)]
Mistake 2:
In the last line, second step,
The 2 in the numerator cannot be cancelled with the 2 in the denominator in this way.
The correction solution:
6/(x-2) – 4(x+2)
= [6(x+2)-4(x-2)]/[(x-2)(x+2)] <-- (no change)
= (6x + 12 – 4x + 8)/[(x-2)(x+2)]
= (2x + 20) / [(x-2)(x+2)]
= [2(x+10)] / [(x-2)(x+2)]

2007-12-10 11:58:55 補充：
7.(a)[(ab^2)/(a^2)]x(ab)=[(b^2)/(a)]x(ab)=b^38.(a)(a^3b^2)x√(a^4/b^6)=(a^3b^2)x(a^2/b^3)=a^(3+2) b^(2-3)=a^5 b^(-1)=(a^5)/(b)

2007-12-11 00:46:59 補充：
Sorry, 因為有事遲了完成.5. ∠CAB = 180º - 120º (adj. ∠s on st. line)In ΔACB,∠ACB + ∠CAB + ∠CBA = 180º (∠ sum of Δ)∠ACB + 56º + 60º = 180º∠ACB = 64º∠ADB = ∠ACB (∠s in the same segment)∴ ∠ADB = 64º

2007-12-11 00:48:04 補充：
6. OJ = OK (radius)∠OJK = ∠OKJ = 22º (base ∠s, isos. Δ)In ΔOJK,∠OJK + ∠OKJ + ∠JOK = 180º (∠ sum of Δ)∠JOK + 22º + 22º = 180º∠JOK = 136º∠OJL = 90º (tangent perpendicular to radius)∠OKL = 90º (tangent perpendicular to radius)to be continued...

2007-12-11 00:48:22 補充：
∠OJL + ∠OKL + ∠JOK + ∠JLK = 360º (∠ sum of polygon)90º + 90º + 136º + ∠JLK = 360º∴∠JLK = 44º