✔ 最佳答案
question 1:
(1): y = 2x - 5 ...... (3)
Put (3) into (2):
x2 + x(2x - 5) = 2
x2 + 2x2 - 5x = 2
3x2 - 5x - 2 = 0
(3x + 1)(x - 2) = 0
x = -1/3 or x = 2
Put the value of x into (3)
When x = -1/3, y = 2(-1/3) - 5 = -17/3
When x = 2, y = 2(2) - 5 = -1
Ans: x = -1/3, y = -17/3 or x = 2, y = -1
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question 2:
(e3x + e-3x)/(ex + e-x)
= [e3x(e3x + e-3x)]/[e3x(ex + e-x)]
= (e3xe3x + e3xe-3x)/(e3xex + e3xe-x)
= (e3x+3x + e3x-3x)/(e3x+x + e2x)
= (e6x + 1)/(e4x + e2x)
= [(e2x)3 + 1]/[(e2x)2 + (e2x)]
= [(5)3 + 1]/[(5)2 + 5]
= [125 + 1]/[25 + 5]
= 126/30
= 21/5
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question 3 :
Let n and (n + 1) be the two consecutive positive integers.
n(n + 1) + [n + (n + 1)] = 71
n2 + n + n + n + 1 = 71
n2 + 3n - 70 = 0
(n - 7)(n + 10) = 0
n = 7 or n = -10 (rejected)
n + 1 = 8
Ans: The two integers are 7 and 8.
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question 4:
(a)
10x2 + 4x + 1 = 2ax ( 2 - x)
10x2 + 4x + 1 = 4ax - 2ax2
(10 + 2a)x2 + (4 - 4a)x + 1 = 0
The equation has real roots. ∆ ≥ 0
(4 - 4a)2 - 4(10 + 2a) ≥ 0
16 - 32a + 16a2 - 40 - 8a ≥ 0
16a2 - 40a - 24 ≥ 0
2a2 - 5a - 3 ≥ 0
(2a + 1)(a - 3) ≥ 0
a ≤ -1/2 or a ≥ 3
(b)
The equation has repeated roots. ∆ = 0
(4 - 4a)2 - 4(10 + 2a) = 0
(2a + 1)(a - 3) = 0
a = -1/2 or a = 3
(c)
The equation has no real roots. ∆ < 0
(4 - 4a)2 - 4(10 + 2a) < 0
(2a + 1)(a - 3) < 0
-1/2 < a < 3
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question 5:
(1): x = 3 - y ...... (3)
Put (3) into (2):
33-y + 3y = 12
(33/3y) + 3y = 12
Multiply each term of the both sides by 3y:
33 + (3y)2 = 12(3y)
(3y)2 - 12(3y) + 27 = 0
(3y - 9)(3y - 3) = 0
3y = 9 or 3y = 3
3y = 32 or 3y = 31
y = 2 or y = 1
Put the value of y into (3):
When y = 2, x = 1
When y = 1, x = 2
Ans: x = 1, y = 2 or x = 2, y = 1
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