数学题一坨1.x^4+mx^2+
1、x^4 + mx^2 + nx - 16含有因式(x-1)和(x-2)，则m =_____，n =_____。
解:将x = 1，x = 2分别代入x^4 + mx^2 + nx - 16得
1 + m + n - 16 = 0 → m + n = 15
16 + 4m + 2n - 16 = 0 → 2m + n = 0
即 m = -15，n = 30
2、化简求值(2x^2 - 6x + 5)^2 - (2x^2 - 9x + 4)^2 - (2x - 3)^2，其中x = -1。
解:
(2x^2 - 6x + 5)^2 - (2x^2 - 9x + 4)^2 - (2...全部
1、x^4 + mx^2 + nx - 16含有因式(x-1)和(x-2)，则m =_____，n =_____。
解:将x = 1，x = 2分别代入x^4 + mx^2 + nx - 16得
1 + m + n - 16 = 0 → m + n = 15
16 + 4m + 2n - 16 = 0 → 2m + n = 0
即 m = -15，n = 30
2、化简求值(2x^2 - 6x + 5)^2 - (2x^2 - 9x + 4)^2 - (2x - 3)^2，其中x = -1。
解:
(2x^2 - 6x + 5)^2 - (2x^2 - 9x + 4)^2 - (2x - 3)^2
= [(2x^2 - 6x + 5) + (2x - 3)][(2x^2 - 6x + 5) - (2x - 3)] - (2x^2 - 9x + 4)^2
= (2x^2 - 4x + 2)(2x^2 - 8x + 8) - (2x^2 - 9x + 4)^2
= [2(x - 1)(x - 2)]^2 - (2x^2 - 9x + 4)^2
= (2x^2 - 6x + 4 + 2x^2 - 9x + 4)(2x^2 - 6x + 4 - 2x^2 + 9x - 4)
= (4x^2 - 15x + 8)(3x)
当x = -1时，
原式 = 4x^2 - 15x + 8)(3x) = [4×(-1)^2 - 15×(-1) + 8][3×(-1)] = -81
3、若abc = 1,则a/(ab + a + 1) + b/(bc + b + 1) + c/(ac + c + 1)的值是_____。
4、已知a = 2 + √3,b = 2 - √3,求 [a + √(ab)]/[√(ab) + b] + [√(ab) - b]/[a -√(ab)] 的值。
解:
[a + √(ab)]/[√(ab) - b] + [√(ab) - b]/[a -√(ab)]
= {[a + √(ab)][a -√(ab)] + [√(ab) - b][√(ab) - b]}/{[√(ab) - b][a -√(ab)]}
= [a^2 - ab + ab - 2b√(ab) + b^2]/[a√(ab) - ab - ab + b√(ab)]
= [(a + b)^2 - 2ab - 2b√(ab)]/[√(ab)(a + b) - 2ab]
= {(a + b)^2 - 2√(ab)[√(ab) - b]}/{√(ab)[(a + b) - 2√(ab)]}
将 a + b = 2 + √3 + 2 - √3 = 4，ab = (2 + √3)(2 - √3) = 1代入上式得
原式 = [16 - 2/(1 - 2 + √3)]/(4 - 2)
= 8 - 1/(√3 - 1)
= 8 - (√3 + 1)/2
= (1/2)(15 - √3)
5、已知x^2 - 3x + 1 = 0,求x^2 + 1/x^2的值。
解:
x1、2 = (3 ± √5)/2
当 x = (3 +√5)/2时，
x^2 + 1/x^2
= 3x + 1 + 1/(3x + 1)
= (3/2)(3 +√5) + 1 + 1/[(3/2)(3 +√5)]
= 5 + (4/3)√5
当 x = (3 -√5)/2时
x^2 + 1/x^2
= 3x - 1 + 1/(3x - 1)
= (3/2)(3 -√5) + 1 + 1/[(3/2)(3 -√5)]
= 6 - (1/2)√5
。
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