115 香山高中_國中部

[LS0722]

請參閱附件，另外請教4、8、11、21、25的做法，感謝！

[thepiano]

第 4 題
a_1b_1 = 20
a_2b_2 = (a_1 + d)b_1r = 20r + db_1r = 19
db_1r = 19 - 20r

a_3b_3 = (a_1 + 2d)b_1r^2 = 20r^2 + 2db_1r^2 = 14
db_1r^2 = 7 - 10r^2

db_1r^2 = db_1r * r = (19 - 20r)r = 7 - 10r^2
10r^2 - 19r + 7 = 0
r = 1/2 或 7/5

a_4b_4 = (a_1 + 3d)b_1r^3 = 20r^3 + 3db_1r^3 = 20r^3 + 3db_1r^2 * r = 20r^3 + 3(7 - 10r^2)r = -10r^3 + 21r = 37/4 或 49/25


第 8 題
a^2 + b^2 = c^2
(a/c)^2 + (b/c)^2 = 1
x^2 + y^2 = 1
(x + y)^2 - 2xy = 1
xy = [(x + y)^2 - 1] / 2

13xy = 15(x + y) - 15
13[(x + y)^2 - 1] / 2 = 15(x + y) - 15
13(x + y)^2 - 30(x + y) + 17 = 0
x + y = 17/13 或 1 (不合，因 a + b ≠ c)


第 11 題
a / (b + c) + b / (c + a) + c / (a + b) = 1
(a + b + c)[a / (b + c) + b / (c + a) + c / (a + b)] = a + b + c
[a^2 / (b + c)] + a + [b^2 / (c + a)] + b + [c^2 / (a + b)] + c = a + b + c
a^2 / (b + c) + b^2 / (c + a) + c^2 / (a + b) = 0


第 21 題
3 - a ≠ 0，a ≠ 3

|a| - 3 ≧ 0，|a| ≧ 3
3 - |a| ≧ 0，|a| ≦ 3
|a| = 3

故 a = -3

A = [(-6) / (-2)]^200 = 3^200 ≡ 1 (mod 10)


第 25 題
x^2 + ax + 2b = 0 的解為實數，a^2 - 8b ≧ 0，b ≦ a^2/8
x^2 + 2bx + a = 0 的解為實數，(2b)^2 - 4a ≧ 0，b^2 ≧ a，b ≧ √a
√a ≦ b ≦ a^2/8
√a ≦ a^2/8
a ≦ a^4/64
a ≧ 4

a + b ≧ a + √a = 4 + 2 = 6