重酬!!! EASY easy MATHS (new)

1)Sub y = 2x+1 into (x+k)² + y² = 5 ,(x² + 2kx + k²) + (4x² + 4x + 1) = 55x² + (2k+4)x + k² - 4 = 0△ = (2k+4)² - 4*5(k² - 4) = 0(k+2)² - 5(k² - 4) = 0(k + 2)² - 5(k - 2)(k + 2) = 0(k + 2)(k + 2 - 5(k - 2)) = 0(k + 2)(4k - 12) = 0k = - 2
or k = 32)1/a² + 1/b²= (a² + b²) / (ab)²= [(a+b)² - 2ab] / (ab)²= [(1/2)² - 2(-5/2)] / (-5/2)²= (21/4) / (25/4)= 21/253)log 3^(log x) = (log x) (log 3)and log x^(log 3) = (log 3) (log x) ,So 3^(log x) = x^(log 3) = kk² - 2(k + k) + 3 = 0k² - 4k + 3 = 0(k - 1)(k - 3) = 0k = 1 or k = 33 ^ log x = 1 or 3 ^ log x = 33 ^ log x = 3^0 or 3 ^ log x = 3^1log x = 0 or log x = 1x = 1 or x = 10a + b = 1 + 10 = 11


2010-12-29 20:42:25 補充：
Q2 : You need to change the equation includes a+b and ab only ,
by the relation between roots and coefficients , sub
a+b = - (-1)/2 = 1/2
ab = - 5/2
into the equation to find the value.

2010-12-29 20:42:38 補充：
Q3 :

You need to prove 3^logx = x^log3 first.

Since log (3^logx) = logx * log3 ,
log (x^log3) = log3 * logx .

So we have proved
log (3^logx) = log (x^log3)
i.e. 3^logx = x^log3

2010-12-29 20:44:06 補充：
Then we let k be 3^logx = x^log3 to find the value of k (i.e. the value of 3^logx) by solving the quadratic equation.

k = 1 or 3

i.e. 3^logx = 1 or 3

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Please explain more about Q2 and Q3!!! Wow, you can get 28pts...so many sponsers.