請教amathe高手！！

p= (x^2+2x+8) / (x-2)

p(x - 2) = x^2 + 2x + 8

px - 2p = x^2 + 2x + 8

x^2 + (2 - p)x + (8 + 2p) = 0

since x is real, b^2 - 4ac > = 0

(2 - p)^2 - 4(1)(8 + 2p) > = 0

(p^2 - 4p + 4) - 32 - 8p > = 0

p^2 - 12p - 28 > = 0

(p - 14) (p + 2) > = 0

p > = 14 or p < = -2

so,

(x^2+2X+8)/(x-2) > = 14 or (x^2+2X+8)/(x-2) < = -2


│x^2+2x+8/x-2│可取值的範圍是

│x^2+2x+8/x-2│>=2

2

(i)
A B C=180

(cosA)^2 - (cosB)^2=sin(A B)sin(B-A)
(cosA)^2 - (cosB)^2=sin(180-C)sin(B-180 B C)
(cosA)^2 - (cosB)^2 (sinC)^2=sinCsin(B-180 B C) (sinC)^2

(cosA)^2 - (cosB)^2 (sinC)^2
=sinC[sin(2B C-180) (sinC)]
=2sinC[sin[(2B 2C-180)/2]cos[(2B-180)/2]]
=2sinCsin(B C-90)cos(B-90)
=2sinCcosB(sin(B C))
=2sinCcosB(sin(180-(B C))
=2cosAsinBsinC

(ii)
(cosA)^2 - (cosB)^2 - (cosC)^2 = -1
(cosA)^2 - (cosB)^2 - (cosC)^2 = -(cosC)^2 - (sinC)^2
(cosA)^2 - (cosB)^2 (sinC)^2 = 0
2cosAsinBsinC=0
i.e. cosA or sinB or sinC = 0
i.e. A = 90 or B or C = 0
so ABC is a right-angled triangle

2008-04-13 20:39:33 補充：
第二題有些加號沒有了(沒手機認證)﹐所以請你自己加回

1) p=(x^2+2x+8)/(x-2) => x^2+(2-p)x+(8+2p)=0
Delta>=0
(2-p)^2-4(1)(8+2p)>=0
解完就p<=-2 or p>=14
l(x^2+2x+8)/(x-2)l>=14
2) L.H.S.=(cosA)^2 - (cosB)^2+ (sinC)^2
= sin(A+B)sin(B-A)+(sinC)^2
= sinC[sin(B-A)+sin(A+B)]
= sinC(2sinBcosA)
= 2cosAsinBsinC
= R.H.S.

2008-04-13 11:45:00 補充：
(cosA)^2 - (cosB)^2-(cosC)^2 = -1
(cosA)^2-(cosB)^2+1-(cosC)^2=0
(cosA)^2-(cosB)^2+(sinC)^2=0
2cosAsinBsinC=0
cosA=0 => A=90*
偶然路過, 本來都想答, 不過......既然喺你眼中中一嘅都比我更算得上係高手嘅話, 答嚟陣哂gas = =