請幫我解6題國中多項式暨平方根的數學,謝謝?

2015-10-10 6:19 am

回答 (1)

2015-10-10 9:06 am
1.
分別把四數平方:
(5/3)² = 25/9
[√(5/3)]² = 5/3 = 15/9
[5/√3]² = 25/3 = 75/9
[(√5)/3]² = 5/9

四數的平方中,以 [(√5)/3]² = 5/9 為最小。
因此,四數中以 (√5)/3 為最小。 ...... 答案選(D)


2.
2^48 - 1
= (2^24)^1 - 1
= (2^24 + 1)(2^24 - 1)
= (2^24 + 1)[(2^12)^2 - 1]
= (2^24 + 1)(2^12 + 1)(2^12 - 1)
= (2^24 + 1)(2^12 + 1)[(2^6)^2 - 1]
= (2^24 + 1)(2^12 + 1)(2^6 + 1)(2^6 - 1)
= (2^24 + 1)(2^12 + 1)(64 + 1)(64 - 1)
= (2^24 + 1)(2^12 + 1) × 65 × 63

此兩數為 63 及 65。
較小者 = 63


3.
由圖線可知:
(1) a < 0 所以 a = -|a|
(2) b > 0 所以 b = |b|

√a² - |a - b|
= |a| - |-|a| - |b||
= |a| - |-(|a| + |b|)|
= |a| - ||a| + |b||
= |a| - |a| - |b|
= -|b|
= -b


4.
如下圖把圖形分割成兩個直角三角形。

圖形總面積
= 兩個直角三角形面積的總和
= (1/2)(3x + 5 + 5)(4x + 2) + (1/2)(2x - 4)(2 + 5x - 2)
= 11x² + 13x + 10

兩個沒斜線的直角三角形面積總和
= (1/2)(5)(4x + 2) + (1/2)(2)(2x - 4)
= 12x + 1

斜線部分的面積
= 圖形總面積 - 兩個沒斜線的直角三角形面積總和
= (11x² + 13x + 10) - (12x + 1)
= 11x² + x + 9


5.
當 0 < x < 1 時:
由於 1/x > x,故此 x - (1/x) < 0

x² + (1/x²) = 27
x² - 2x(1/x) + (1/x)² = 27 - 2x(1/x)
[x - (1/x)]² = 25
x - (1/x) = 5 (不合、因 x - (1/x) < 0) 或 x - (1/x) = -5

所以,x - (1/x) = -5


6.
108 = 2^2 × 3^3

當 a = 3 時: √(108 × a) = √(108 × 3) = √(2^2 × 3^4) = 18
當 b = 3 時: √(108 / b) = √(108 / 3) = √(2^2 × 3^2) = 6
當 c = 13 時: √(108 + c) = √(108 + 13) = √121 = 11
當 d = 8 時: √(108 - d) = √(108 - 8) = √100 = 10

(a + b + c + d + 9) 的平方根
= ±√(a + b + c + d + 9)
= ±√(3 + 3 + 13 + 8 + 9)
= ±√36
= ±6


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