高中職數學4題求下列四題怎麼解(在詳細解說內) 要有解法 (如果打字不好表達 可手寫拍照 FB或是LINE給我?

8!/2!2!2!2!=7! =5040 ?沒錯吧
第三題有想過一種排法
ccdd亂數排列*5個空格取2格*3個空格取2格(ccdd穿插aabb且aabb4個字母皆不相鄰)+ccdd亂數排列*5個空格取1格*4個空格取2格(ccdd穿插aabb且aa再一起bb分開放)+ccdd亂數排列*5個空格取1格*4個空格取2格(ccdd穿插aabb且bb再一起aa分開放)+ccdd亂數排列*5個空格取1格*4個空格取1格(ccdd穿插aabb且aa再一起bb也在一起)
[4!/(2!2!)]*(C5取2)*(C3取2)+[4!/(2!2!)]*(C5取1)*(C4取2)+[4!/(2!2!)]*(C5取1)*(C4取2)++[4!/(2!2!)]*(C5取1)*(C4取1)
=6*10*3+6*5*6+6*5*6+6*5*4=660
不知道這想法有沒有正確
前天解出來了～其他4題跟自己算的有一樣

1.
已知 in any triangle with the usual dimension a,b,c and opposing angles A,B,C,
T=2*area=(bc)sin(A)=(ca)sin(B)=(ab)sin(C)
=>
(c)sin(B)=T/a=(b)sin(C)..................(1a)
(a)sin(C)=T/b=(c)sin(A)..................(1b)
(b)sin(A)=T/c=(a)sin(B)..................(1c)
The given determinant expands to
D=97(csin(B)-bsin(C))+98(asin(C)-csin(A))+99(bsin(A)-asin(B))
用 (1a),(1b),(1c)
D=97(T/a-T/a)+98(T/b-T/b)+99(T/c-T/c)=0

2.
f(x)-f(-1)
a=lim -----------
x->-1 x-(-1)
是 f(x) 導數的定義, 或 f'(-1).

已知 f(x)=(x+1)(x+2)...(x+100), use product rule to find f'(x).
設 g(x)=(x+2)(x+3)...(x+100)
f(x)=(x+1)g(x)=(x+1)'g(x)+(x+1)g(x)=g(x)+(x+1)g'(x)
=>
f(-1)=g(x)+(-1+1)g'(x)=g(x)=(x+2)(x+3)....(x+100)

3.
Number of distinct arrangements (with 4 pairs of identical letters)
=8!/(2!2!2!2!)=2520
less all those with "ab" or "ba":
Arrangements with "ab"=7!/(2!2!)=1260
Arrangements with "ba"=7!/(2!2!)=1260
plus overcounting, using inclusion/exclusion principle:
arrangements with "ab" and "ab"=6!/(2!2!2!)=90
arrangements with "ab" and "ba"=6!/(2!2!)=180
arrangements with "ba" and "ba"=6!/(2!2!2!)=90
arrangements with "ba" and "ab"=6!/(2!2!)=180

Number of arrangements without "ab" or "ba"
=2520-(1260+1260)+2(90+180)=540


4.
已知 x=3sec(t)+1, y=4tan(t)-2
用恆等式 1+tan(t)^2=sec(t)^2 to eliminate t.
(x-1)^2/3^2-(y+2)^2/4^2=1 為雙曲線, 中心在 (1,-2).
Since one vertex is at (4,-2), with a=3, b=4.
正焦弦長( latus rectum) = 2b^2/a=2*4^2/3=32/3
查核:
焦距: sqrt(3^2+4^2)=5
在 x=5, y=+/-sqrt(16*(5^2/3^2-1))=+/-(4sqrt(16/9))=+/-(16/3)
正焦弦長 (latus rectum) = 2(16/3)=32/3 as before.