114 新北市國中
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Re: 114 新北市國中
第 30 題
以下"度"省略
利用 cos40 + cos80 + cos160 = 0
(cos80)^2 + (cos160)^2 + cos80cos160
= (cos80 + cos160)^2 - cos80cos160
= (-cos40)^2 - (1/2)(cos240 + cos80)
= (cos40)^2 + 1/4 - (1/2)[2(cos40)^2 - 1]
= 3/4
第 33 題
f(x) = (ax + b)q(x) + r
(A) f(x) = (x + b/a)aq(x) + r,商應為 aq(x)
(B)(C) xf(x) = (ax + b)xq(x) + xr = (ax + b)xq(x) + (ax + b)(r/a) - br/a
= (ax + b)[xq(x) + (r/a)] - br/a
第 34 題
設 △DEF = x,CE/AE = a/b
a/b = 32/x
AF/BF = △AFE/△BDF = 5/3
15/(9 + 15 + 32 + x) = △AFE/△ABC = (AF * AE)/(AB * AC) = (5b)/[8(a + b)]
15/(x + 56) = (5b)/[8(a + b)]
x = (32b)/a 代入 15/(x + 56) = (5b)/[8(a + b)]
15/[(32b)/a + 56] = (5b)/[8(a + b)]
化簡可得 a = 2b,x = 16
所求 = 16/72 = 2/9
第 35 題
(1) x = y
代入 x^3 + y^3 = 5(x + y)
可得 2x^3 = 10x
x = 0,√5,-√5
(2) x = -y
代入 x^3 - y^3 = 7(x - y)
可得 2x^3 = 14x
x = 0,√7,-√7
(3) x ≠ y 且 x ≠ -y
x^2 + xy + y^2 = 7
x^2 - xy + y^2 = 5
x = √2 + 1,-√2 - 1,√2 - 1,-√2 + 1
計 9 組解
第 38 題
x^2 + ax + b = 0 有兩實根
判別式 a^2 - 4b ≧ 0,a^2 ≧ 4b
α^2 + β^2 = (α + β)^2 - 2αβ = (-a)^2 - 2b ≧ 9
a^2 ≧ 2b + 9
(a,b) = (4,1)、(4,2)、(4,3)
(5,1)、(5,2)、(5,3)、(5,4)、(5,5)、(5,6)
(6,1)、(6,2)、(6,3)、(6,4)、(6,5)、(6,6)
所求 = 15/6^2 = 5/12
以下"度"省略
利用 cos40 + cos80 + cos160 = 0
(cos80)^2 + (cos160)^2 + cos80cos160
= (cos80 + cos160)^2 - cos80cos160
= (-cos40)^2 - (1/2)(cos240 + cos80)
= (cos40)^2 + 1/4 - (1/2)[2(cos40)^2 - 1]
= 3/4
第 33 題
f(x) = (ax + b)q(x) + r
(A) f(x) = (x + b/a)aq(x) + r,商應為 aq(x)
(B)(C) xf(x) = (ax + b)xq(x) + xr = (ax + b)xq(x) + (ax + b)(r/a) - br/a
= (ax + b)[xq(x) + (r/a)] - br/a
第 34 題
設 △DEF = x,CE/AE = a/b
a/b = 32/x
AF/BF = △AFE/△BDF = 5/3
15/(9 + 15 + 32 + x) = △AFE/△ABC = (AF * AE)/(AB * AC) = (5b)/[8(a + b)]
15/(x + 56) = (5b)/[8(a + b)]
x = (32b)/a 代入 15/(x + 56) = (5b)/[8(a + b)]
15/[(32b)/a + 56] = (5b)/[8(a + b)]
化簡可得 a = 2b,x = 16
所求 = 16/72 = 2/9
第 35 題
(1) x = y
代入 x^3 + y^3 = 5(x + y)
可得 2x^3 = 10x
x = 0,√5,-√5
(2) x = -y
代入 x^3 - y^3 = 7(x - y)
可得 2x^3 = 14x
x = 0,√7,-√7
(3) x ≠ y 且 x ≠ -y
x^2 + xy + y^2 = 7
x^2 - xy + y^2 = 5
x = √2 + 1,-√2 - 1,√2 - 1,-√2 + 1
計 9 組解
第 38 題
x^2 + ax + b = 0 有兩實根
判別式 a^2 - 4b ≧ 0,a^2 ≧ 4b
α^2 + β^2 = (α + β)^2 - 2αβ = (-a)^2 - 2b ≧ 9
a^2 ≧ 2b + 9
(a,b) = (4,1)、(4,2)、(4,3)
(5,1)、(5,2)、(5,3)、(5,4)、(5,5)、(5,6)
(6,1)、(6,2)、(6,3)、(6,4)、(6,5)、(6,6)
所求 = 15/6^2 = 5/12