數學題求救要詳細步驟

i)
6cos²x+6sin²x
=6(cos²x+sin²x)
=6(1)
=6

ii)
(1+csc x)/csc x
=1/csc x+csc x/csc x
=sin x+1

(A)
domain: all real numbers
range: f(x)≥2/3

(B)
domain: x≠0
range: g(x)≠0

2014-10-05 18:47:50 補充：
只是值域、範圍等。
难以解釋。

(A) f(x)=3x^2-2x+1
= 3(x^2 - 2/3 x + 1/3)
= 3(x^2 - 2/3 x + 1/9 +2/9)
= 3[(x - 1/3)^2 + 2/9]
= 3(x - 1/3)^2 +2/3
Since 3(x - 1/3)^2 ≥ 0, f(x) ≥ 2/3.
Hint: You need to convert the equation into another form which can show its characteristics (e.g. maximum/minimum values).

(B) g(x)=1/x
When x tends to zero, g(x) tends to infinity.
However, g(x) can never reach infinitiy.
Thus, domain: x≠0

When x tends to infinity (no matter positive or negative), g(x) tends to zero.
However, x can never reach infinity.
Thus, range: g(x)≠0

我本來想答你，但見你好像不是太明白個概念，所以先搞清楚一下。

你要先明白什麼是 domain 和 range，與其你在等待，不如先翻開書本看看，你指出你不明白的地方，我們可以講解。