F.4 MATHS..

1..已知x和y都是正數且y的一部分是固定不變的,而另一部分隨x^2反變,
當x = 5時,y = 8;又當x = 3時,y = 28
a)試以x表y
y = ｜b + a/ x^2｜
8 = b + a/25 ………..(1)
28 = b + a/9………..(2)
(2) - (1)
20 = a( 1/9 – 1/25 )
= a(16/225)
a = 1125/4
根據(2)
b = 28 – 1125/36
= -117/36
所以,
y = ｜-117/36 + 1125/ 4x^2｜

b)當x = 4時,求y的值
當x = 4時
y = ｜-117/36 + 1125/ 4(4)^2｜
= 14.33

2..已知y的一部分是常數,而另一部分隨x反變,當x = 6時,y = 7;又當x = 4時,y = 8
a)試以x表y
y = b + a/ x
7 = b + a/6 ………..(1)
8 = b + a/4 ………..(2)
(2) - (1)
1= a/4 - a/6
a =12
根據(2)
8 = b + 12/4
b = 5
所以,
y = 5 + 12/x

b)當x = 3時,求y的值
當x = 3時,
y = 5 + 12/3
=9


3..己知y的一部分隨x正變,而另一部分則隨x^2正變,當x = 1時,y = 1'又當x = 2時,y = 0.
當x = 3時,求y的值..
y = ax^2 +b x
1 = a + b ……………(1)
0 = 4a + 2b……………(2)
2*(1) – (2),
-2a = 2
a = -1
根據(1)
1 = -1+b
b = 2
所以,
y = -x^2 + 2x
or y = x( 2 – x )

當x = 3時,
y = 3( 2 – 3 )
= -3

1..已知x和y都是正數且y的一部分是固定不變的,而另一部分隨x^2反變,
當x = 5時,y = 8;又當x = 3時,y = 28
a)試以x表y
y = ｜b + a/ x^2｜
8 = b + a/25 ………..(1)
28 = b + a/9………..(2)
(2) - (1)
20 = a( 1/9 – 1/25 )
= a(16/225)
a = 1125/4
根據(2)
b = 28 – 1125/36
= -117/36
所以,
y = ｜-117/36 + 1125/ 4x^2｜

b)當x = 4時,求y的值
當x = 4時
y = ｜-117/36 + 1125/ 4(4)^2｜
= 14.33

2..已知y的一部分是常數,而另一部分隨x反變,當x = 6時,y = 7;又當x = 4時,y = 8
a)試以x表y
y = b + a/ x
7 = b + a/6 ………..(1)
8 = b + a/4 ………..(2)
(2) - (1)
1= a/4 - a/6
a =12
根據(2)
8 = b + 12/4
b = 5
所以,
y = 5 + 12/x

b)當x = 3時,求y的值
當x = 3時,
y = 5 + 12/3
=9


3..己知y的一部分隨x正變,而另一部分則隨x^2正變,當x = 1時,y = 1'又當x = 2時,y = 0.
當x = 3時,求y的值..
y = ax^2 +b x
1 = a + b ……………(1)
0 = 4a + 2b……………(2)
2*(1) – (2),
-2a = 2
a = -1
根據(1)
1 = -1+b
b = 2
所以,
y = -x^2 + 2x
or y = x( 2 – x )

當x = 3時,
y = 3( 2 – 3 )
= -3

哈哈=_=" 功課要自己做 XD