Page 21 - ePD310_技術高中數學C第二冊學習講義含解析本_課本PDF

Chapter 1 式的運算 15

1-1 實力測驗 解析 P.6、7

1. ɨΐО٫ʔ݊ x ٙεධόkc
x 2
2
2
(A) 2(x + 2)(x - 3)c(B) x + + 1c(C) x + - 3c(D)5f
2 x
4
3
2. ɨΐО٫މ x ٙεධόkc(A) x + c(B) 5c(C) |x - 2|c(D) f
3x + 1

2
3
3. ணεධό f (x) = 5x - 2x - x + 4 ٙϣᅰމ adධᅰމ bd࠯ධڷᅰމ cdۆ
a + b + c = kc(A) 12c(B) 6c(C) 14c(D) 7f

4. ʊٝՇεධό f (x) ձ g(x) ٙϣᅰʱйމ 3 ϣձ 4 ϣdۆ f (x) × g(x) ٙϣᅰމjc

(A) 3c(B) 4c(C) 7c(D) 12f

2
2
3
5. Ӌ 2x + x - 4 ৰ˸ x + 2x - 1 ٙਠόމjc
(A) 2x - 3c(B) 2x - 1c(C) 2x + 1c(D) 2x + 3f

3
3
2
6. ண x + x + 2 = a(x - 1) + b(x - 1) + c(x - 1) + ddۆɨΐО٫މॆkc
(A) a = 0c(B) b = -3c(C) c = 4c(D) d = -4f

2
7. ߰ f (x) = 2x - 7dg(x) = 3x - 4 ˲ 3f (x) - 2x × g(x) = h(x)dۆjc
(A) h(x) = 8x + 21c(B) h(0) = 21c(C) h(1) = -13c(D) h(1) = 13f

2
4
3
3
2
4
8. ண x + 3x - 2x - 3x + 6 = a(x + 1) + b(x + 1) + c(x + 1) + d(x + 1) + ed ۆjc
(A) b + d = ec(B) b + c = dc(C) a + b + c + d + e = 5c(D) b + d = a + ef
1 3
9. ʃ׼Աኽሙ͉ၝΥৰجࣣٙᄳ˙όdӋ੻ f (x) ৰ˸ x + ٙਠόމ q(x)d
2 4
ቱόމ (-5)dՉЪجνɨהΐdۆਠό q(x) ʕdx ධٙڷᅰމj
3
2-1-8-8 -
2
-3+6+3
2-4-2-5

(A) -2c(B) -4c(C) -8c(D) -16f

10. ண (x + ax + 3)(x + b) = x + 7x + cx + 6dۆ a + b + c ʘ࠽މjc
2
3
2
(A) 20c(B) 18c(C) 2c(D) 1f