請幫忙2題數學題目

1.
[2t - (1/2t)]^10

t² 項 = 10Cr(2t)^(10-r) (1/2t)^r
10 - r - r = 2
10 - 2r = 2
2r = 8
r = 4

t^2 之係數
= 10C4 (2)^6 (1/2)^4
= (10!/4!6!) x (2^2)
= 210 x 4
= 840


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2.
情況一：
[(a/x^2) - (√3)x]^6

常數項 = 6Cr(a/x^2)^(6-r) [-(√3)x]^r

(1/x^2)^(6 - r) x^r = 1
r - 2(6 - r) = 0
r - 12 + 2r = 0
r = 4

6C4 (a/x^2)^2 [-(√3)x]^4 = 720
(6!/4!2!) (a^2 / x^4) (9x^4) = 720
135 a^2 = 720
a =(4√3)/3 或 a = -(4√3)/3

情況二：
[(a/x^2) -√(3x)]^6

常數項 = 6Cr(a/x^2)^(6-r) [-√(3x)]^r

(1/x^2)^(6 - r) x^(r/2) = 1
(r/2) - 2(6 - r) = 0
(r/2) - 12 + 2r = 0
5r/2 = 12
r = 4.8 (不合)

到下面的網址看看吧

▶▶http://misshare168.pixnet.net/blog/post/86950298

有一個人不去回答問題，只說自己是應數碩士，你會相信這個人嗎？

((2t- 1/(2t))^2)*10 ?

(((a/x^2)-3x^(1/2))^2)*6


題目都不講清楚怎麼解