form 4 math(急!!)

1)
2+ √(x-1) = √(5x-9)
4 + 4√(x-1) + (x-1) = (5x-9)
4√(x-1) = 5x-9 + 1-x - 4
4√(x-1) = 4x - 12
√(x-1) = x - 3
x - 1 = x^2 - 6x + 9
x^2 - 7x + 10 = 0
(x - 2)(x - 5) = 0
x = 2 or x = 5 ,
When x = 2 , LHS = 3 not = RHS = 1 , rejected
Hence x = 5 (LHS = 4 = RHS)
2)
(1/2log256+2logx^2+8logy)(log(2x)+4log√y)
= (log (256 ^ 1/2) + logx^4 + logy^8)(log(2x) + log (√y)^2)
= (log16 + logx^4 + logy^8) / (log (2x) + log y^2)
= (log [16(x^4)(y^8)]) / (log 2xy^2)
= ( log (2xy^2)^4 ) / (log 2xy^2)
= 4 log (2xy^2) / (log 2xy^2)
= 4
3)
x^2-5mx+6m^2=0
(x - 2m)(x - 3m) = 0
x = 2m
or
x = 3m

2010-06-08 23:50:13 補充：
Some mistake :

for Q2 :

(log (256 ^ 1/2) + logx^4 + logy^8)(log(2x) + log (√y)^2)

should be

(log (256 ^ 1/2) + logx^4 + logy^8)(log(2x) + log (√y)^4)

very very very very ... ... GOOD !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!