数学Paper 1讨论专区

Log

回复 2570# 天空之道
from B,
P((AnB)|(AUB))= (1-x)/x
since  0 ≤ P((AnB)|(AUB)) ≤ 1
         0 ≤ （1-x)/x ≤ 1
solve the inequality , u ll show  1/2 ≤ x ≤ 1

Log

回复 2573# huatiang
rational number is a number which can be express in term of m/n form, which m and n are integers , n ≠ 0.
1= 5/5, 1= 100001/100001 etc..

peaceboy

回复 2575# jose


   题目怪怪的 哪里来的？不像是stpm syllabus

jose

回复  jose


   题目怪怪的 哪里来的？不像是stpm syllabus
peaceboy 发表于 21-5-2011 08:52 PM

   老师好像是从pelangi的书里抄出来的

TianSang

回复 2578# peaceboy


    谢啦 xD

given that 2y=a^(x) + a^(-x) = 2y, where a>1 and x>0, prove that a^(x) = y + Square root of ( y^(2) - 1 )
similarly, if 2z = a^(3x) + a^(-3x), prove that z = 4y^(3) -3y

peaceboy

回复 2585# TianSang


   given that 2y=a^(x) + a^(-x) = 2y, where a>1 and x>0, prove that a^(x) = y + Square root of ( y^(2) - 1 )
similarly, if 2z = a^(3x) + a^(-3x), prove that z = 4y^(3) -3y

a^(x) + a^(-x) = 2y
(a^2x + 1) /a^x = 2y
(a^2x + 1) = 2ya^x
a^2x - 2ya^x = -1
a^2x - 2ya^x +y^2 = y^2 -1
(a^x - y) =  Square root of ( y^(2) - 1 )
a^(x) = y + Square root of ( y^(2) - 1 )




2y=a^(x) + a^(-x) ----1
2z = a^(3x) + a^(-3x)----2


2/1
z/y = [(a^6x +1)/ a^3x] / [(a^2x+1)/a^x]
     =  (a^6x +1)(a^x) /  (a^3x)(a^2x+1)
      =(a^6x +1) /  (a^2x)(a^2x+1)
      = (a^2x +1)(a^4x - a^2x + 1) /  (a^2x)(a^2x+1)
      = (a^4x - a^2x + 1) /  (a^2x)


a^2 = 2y^2-1 + 2y √ (y^2 -1)
a^4 = 4y^4 - 4y^2 +1 + 4y^4 - 4y^2  + (4y^2 - 2)(2y) √ (y^2 -1)
       = 8y^4 -8y^2 +1 + 8y^3√ (y^2 -1) - 4y √ (y^2 -1)
      
a^4- a^2x + 1 =  8y^4 -8y^2 +1 + 8y^3√ (y^2 -1) - 4y √ (y^2 -1) - 2y^2+1 - 2y √ (y^2 -1) +1
                      =8y^4 -10y^2 +3+ 8y^3√ (y^2 -1)- 6y √ (y^2 -1)

=(4y^2-3)(2y^2 -1)+  (4y^2-3)[2y √ (y^2 -1)]
                       = (4y^2-3)(2y^2 -1+2y √ (y^2 -1)





z/y =
(a^4x - a^2x + 1) /  (a^2x)
         =(4y^2-3)(2y^2 -1+2y √ (y^2 -1) / (2y^2 -1+2y √ (y^2 -1)
         = (4y^2-3)
z=4y^(3) -3y

huatiang

回复 2582# Log


  Can m=n? 怪怪的...

TianSang

回复 2586# peaceboy


    thankiu thankiu 对不起呀小弟很笨 =X

given that z = x + yi and w = z + 8i over z - 6, z nt equal to 6. If w is totally imaginary, show that x^(2) + y ^(2) + 2x - 48 = 0

peaceboy

回复 2588# TianSang

确定题目没有错？我最多show到x^2 + y^2 -6x + 8y = 0 罢鸟

TianSang

回复 2589# peaceboy


    given that z = x + yi and w = (z + 8i) over (z - 6), z nt equal to 6. If w is totally imaginary, show that x^(2) + y ^(2) + 2x - 48 = 0

就这样 O.O

Allmaths

回复  peaceboy


    given that z = x + yi and w = (z + 8i) over (z - 6), z nt equal to 6. If w  ...
TianSang 发表于 22-5-2011 11:45 PM

   

题目错了啦。。。

peaceboy做对了。。。

这个题目是从pelangi来的。。。看了几百次。。。

Allmaths

回复  peaceboy


    given that z = x + yi and w = (z + 8i) over (z - 6), z nt equal to 6. If w  ...
TianSang 发表于 22-5-2011 11:45 PM

   
话说是 (z+8i)/(z-6) 还是 (z+8i)/(z-6i)?

TianSang

回复 2592# Allmaths


    还是 w = ( z+8i ) over ( z-6 ). 可能今年的pelangi印错了吧 >< 可以把算草写出来吗？ 谢啦 ^_^

Log

回复 2587# huatiang
ya...can

一支牙刷

请问下 37+20surd3 怎样做？

chingying

请问下 37+20surd3 怎样做？
一支牙刷 发表于 24-5-2011 08:49 PM

    i also want to know~~!!!! help us solve this pls!!!!!!
<3<3 thank you so much!!!!

數學神童

let surd(37+20surd3)=a + b surd 3
then square both sides and solve it

gerrykhang

请问下 37+20surd3 怎样做？
一支牙刷 发表于 24-5-2011 08:49 PM

1.    multiple with its conjugate 37-20surd20.
2. add with its conjugate 37-20surd20.
3. solve these two simultaneous equation to get it's answer.

blazex

1)solve the equation (sin^-1 X)-(cos^-1 X)=(sin^-1(1-x))

2)without using table or calculators, show that (cos^-1(3/5))-(tan^-1(-3/4))=pi/2.

peaceboy

回复 2599# blazex


   trigo是math T的吧我只想到画三角形来solve