(20points) Algebraic Fractions & Circles

1.8/x + 3x/x+2 = 4
8(x+2)/x(x+2) + 3x(x)/x(x+2) = 4
8x+2/x²+2x + 3x²/x²+2x = 4
8x+2+3x²=4(x²+2x)
-x²+2=0
x²=2
x=+/-√2
2.2/y+1 + 3/2y+3 = 1
2(2y+3)/(y+1)(2y+3) + 3(y+1)/(2y+3)(y+1) = 1
4y+6/2y²+5y+3 + 3y+1/2y²+5y+3 = 1
7y+7 = 2y²+5y+3
2y²-2y-4=0
y²-y-2=0
(y+1)(y-2) = o
y = -1 or y = 2
3.∠ADB + ∠BDE = 180° (adj. ∠s on st. line)
∠ADB = 76°
∠ACB = 2∠ADB (∠ at centre twice ∠ at circumference)
∠ACB = 152°
4.∠QRS + ∠SPQ = 180° (opp. ∠s, cyclic quad.)
∠SPQ = 59°
Consider ΔPSQ,
∠QSP + ∠SPQ + ∠PQS = 180° (∠ sum of Δ)
∠PQS = 75°
∠SPU = ∠PQS = 75° (∠ in alt. segment)
5.a^-2 b² x √(a^4/b^4)
= a^-2 b² x (a^4/b^4)^1/2
= a^-2 b² x a²b^-2
= 1
6.1/3x+2 - 1/3x+5
=3x+5/(3x+2)(3x+5) - 3x+2/(3x+5)(3x+2)
= 3/9x²+21x+10
7.x^5 y³/x²y^4 X x³y²
=x³y^-1 X x³y²
=x^6 y

2007-12-18 10:18:42 補充：
Sorry,第6題唔知點解會有d nbsp,應該係空白既html碼,照寫番個減號就得啦~

2007-12-19 09:34:35 補充：
第7題我睇漏左個負號,所以...x^5 y³/x²y^4 X x³y^-2=x³y^-1 X x³y^-2=x^6/y³

2007-12-19 09:37:52 補充：
暈啦...第6題~1/3x十2 - 1/3x-5=3x-5/(3x十2)(3x-5) - 3x十2/(3x十2)(3x-5)= -7/9x²-9x-10

2007-12-19 09:40:23 補充：
點解搞到我題題都有d錯...第5題~a^-2 b² x √(a^4/b²)=a^-2 a²b²b^-1=b

2007-12-19 09:50:51 補充：
抄完題目都可以睇錯...1同21.8/x 十 3x/x十2 = 48x十16/x²十2x 十 3x²/x²十2x = 43x²十8x十16=4x²十8x-x²十16=0x=十/- 42.2/y十1 十 3/2y十3 = 14y十6/2y²十5y十3 十 3y十3/2y²十5y十3 = 17y十9 = 2y²十5y十32y²-2y-6 = 0y²-y-3 = 0y = (1十/-√1²-4(1)(-3))/2y = (1十/-√13)/2

1(a) 8/x + 3x/(x+2) = 4
8(x+2) + 3x^2 = 4x(x+2) gives 8x + 16 + 3x^2 = 4x^2 + 8x
so x^2 = 16 gives x = 4 or -4

2(d) 2/(y+1) + 3/(2y+3) = 1
2(2y+3) + 3(y+1) = (2y+3)(y+1) gives 4y + 6 + 3y + 3 = 2y^2 + 5y + 3
so 2y^2 - 2y - 6 = 0 or y^2 - y -3 = 0
y = 1/2 +((4(1)(-3)-1^2)/2 = 1/2 + i(13)^(1/2)
or y = 1/2 - i(13)^(1/2) i =(-1)^(1/2)

3(c) Joint DC and extend to E
Angle ADC = Angle BDC = (180 -104)/2 = 38 (angles on equal chord (radius of circle))
Angle DAC = Angle ADC= 38 and Angle DBC = Angle BDC= 38 (angle of iso, triangle)

Angle ECB = Angle BDC + Angle DBC = 38 + 38 = 76 (ext angle of triangle)
Angle ECA = Angle CDA + Angle DAC = 38 + 38 = 76 (ext angle of triangle)

Angle ACB = Angle ECB + Angle ECA = 76 + 76 = 152

4. Angle SRQ + Angle SPQ = 180 (oppo angles , cyclic quad.)
So Angle SPO = 180 - 121 = 59
Angle POs = 180 - 59 - 46 = 75 (angle sum of triangle)
Angle UPS = Angle POS = 75 (ext angle, cyclic quad)

5(b) a^(-2)b^2 x(a^4/b^2)^(1/2)
= a^(-2+4/2)b^(2-2/2)
= (a^0)(b^1) = b

6(c) 1/(3x+2) - 1/(3x-5)
= ((3x-5)-(3x+2))/((3x+2)(3x-5))
=(3x-5-3x-2)((3x+2)(3x-5))
= -7/(3x+2)(3x-5)

7(c) ((x^5)(y^3))/((x^2)(y^4))(x^3)(y^-2)
= x^(5+3-2)y^(3-2-4)
=(x^6)(y^-3)




























bdc

(3) Draw line connecting D and C.
Angle ADB = 180 - 104 = 76
=> Angle CBD = 76 / 2 = 38
=> Angle BCD = 180 - 38 - 38 = 104
=> Angle ACB = (180 - 104) x 2 = 152