A question

=.=....好似全部都有問題....I hope I can help you~

1. k(3k-4)<0
so 0<4/3

reason:we want to find the range when k(3k-4) ' <0 ', so the answer should under the x-axis,since the curve is open UPWARD,and the x-intercept are 0 and 4/3,by plotting the graph of k(3k-4)=0,we find that when k(3k-4) ' <0 ', the range of k is ....0<4/3

p.s. ' <0 '...means you should find the answers under the x-axis
' >0 '...means you should find the answers above the x-axis

2007-09-29 17:01:52 補充：
補充一句...k(3k-4)<0之後就可以直接出答案,唔需要步驟,因為咁樣係考識唔識我o係上面講ge reason

2007-09-29 17:15:19 補充：
=.=..唔係計錯...打漏左ja..=.=0<4/3同埋係a.maths,唔係maths呀~~

2007-09-29 17:17:19 補充：
0 < k < 4/3

2007-09-29 17:19:40 補充：
唔知點解打極都出唔到個< k < ja......原來要隔個space先打到....=.=

This question is really not a difficult question. However, 5 of you still misunderstood the question, the result is unsatisfactory.

k (3k -4 ) < 0
ie.(A) k < 0 & 3k-4 > 0 or (B)k > 0 & 3k-4 < 0
(A) for 3k - 4 > 0, k > 4/3 , but k < 0, (contradiction , ignore)
(B), 3k-4<0, k <4/3 & k > 0,
Hence, answer : 0 < k < 4/3

2007-09-29 17:00:27 補充：
10年前...我都係A...-_- HKCEE計埋加減數...書都唔駛溫

Let k(3k-4)=0
So k=0 or
3k-4=0
k=4/3

So The final answer: 0 < k < 4/3

k(3k-4)<0

3k^2-4k<0

3k^2<4k

k^2<(4/3)k

k<(4/3)

1 k(3k-4)<0
3k^2-4k<0
3k^2<4k
3k<4
k<4/3

1.) k(3k-4)<0

3k^2-4k<0

3k^2<4k

k^2<(4/3)k

k<(4/3)

2007-09-29 17:24:28 補充：
無所謂~最緊要幫到你^^"

k(3k-4)<0
3k^2-4k<0
3k^2<4k
3k<4
k<4/3