小高一數學問題

(1)
(xy - 1)² + (x + y)(xy + 1) + 5xy
= [(xy + 1) - 2]² + (x + y)(xy + 1) + 5xy
= (xy + 1)² - 4(xy + 1) + 4 + (x + y)(xy + 1) + 5xy
= (xy + 1)² + (x + y)(xy + 1) + xy
= (xy + 1 + x)(xy + 1 + y)


(2)
(x²)² + 5x²y² +9(y²)²
= [(x²)² + 6x²y² +9(y²)²] - x²y²
= (x² + 3y²)² - (xy)²
= (x² + 3y² + xy)(x² + 3y² - xy)


(3)
a³b³ + a²b² - 2
= (a³b³ - 1) + (a²b² - 1)
= (ab - 1)(a²b² + ab + 1) + (ab - 1)(ab + 1)
= (ab - 1)[(a²b² + ab + 1) + (ab + 1)]
= (ab - 1)(a²b² + 2ab + 2)


(4)
x(x²-1) - y(y²-1) + xy(x-y)
= x³ - x - y³ + y + xy(x - y)
= (x³ - y³) - (x - y) + xy(x - y)
= (x - y)(x² + xy + y²) - (x - y) + xy(x - y)
= (x - y)[(x² + 2xy + y²) - 1]
= (x - y)(x + y + 1)(x + y - 1)


(5)
若題目為 (x² - y²)² - 8(x² + y² -2) 之誤，則解法如下：
(x² - y²)² - 8(x² + y² -2)
= (x² - y²)² - 8(x² - y²) - 16y² + 16
= [(x² - y²)² - 8(x² - y²) + 16] - 16y²
= [(x² - y²) - 4]² - (4y)²
= (x² - y² - 4 + 4y)(x² - y² - 4 - 4y)
= [x² - (y - 2)²] [x² - (y + 2)²]
= (x + y - 2)(x - y + 2)(x + y + 2)(x - y - 2)


(6)
x³ + 5x² + 9x + 6
= (x³ + 2x²) + (3x² + 6x) + (3x + 6)
= x²(x + 2) + 3x(x + 2) + 3(x + 2)
= (x + 2)(x² + 3x + 3)


(7)
4(x²)² - 13x² + 1 = 0
[4(x²)² - 4x² + 1] - 9x² = 0
(2x² - 1)² - (3x)² = 0
(2x² + 3x - 1)(2x² - 3x - 1) = 0
2x² + 3x - 1 = 0 或 2x² - 3x - 1 = 0
x = (-3+√17)/4 或 x = (-3-√17)/4 或 x = (3+√17)/4 或 x = (3-√17)/4


8.
x² - 5x + 2√(x²-5x+3) = 12
[√(x² - 5x + 3)]² + 2√(x²-5x+3) - 15 = 0
[√(x² - 5x + 3) + 5][√(x² - 5x + 3) - 3] = 0
√(x² - 5x + 3) = - 5(捨去) 或 √(x² - 5x + 3) = 3
x² - 5x + 3 = 9
x² - 5x - 6 = 0
(x - 6)(x + 1) = 0
x = 6 或 x = -1


(9)
(√3 - 1)x² - 4x + (3 + √3) = 0
x = {4 ± √[(-4)² - 4(√3 - 1)(3 + √3)]} / 2(√3 - 1)
x = [2 ± √(4 - 2√3)] / (√3 - 1)
x = [2 ± (√3 - 1)] / (√3 - 1)
x = (√3 + 1)/(√3 - 1) 或 x = (3 - √3)/(√3 - 1)
x = 2 + √3 或 x = √3

(2)(x2)2+5x2y2+9(y2)2
=(x^2+xy+3y^2)(x^2-xy+3y^2)
令u=x^2,v=y^2改寫為(u+3v)^2-(√uv)^2匴好帶回
(3)a3b3+a2b2-2
=(ab)^3-1+(ab)^2-1
=(ab-1)(a^2b^2+ab+1)+(ab+1)(ab-1)
=(ab-1)(a^2b62+ab+1+(ab+1))
=(ab-1)(a^2b^2+2ab+2) (4)x(x2-1)-y(y2-1)+xy(x-y)
=x^3-y^3-(x-y)+xy(x-y)
=(x-y)(x^2+xy+y^2)-(x-y)+xy(x-y)
=(x-y)[(x^2+xy+y^2)-1+xy]
=(x-y)(x^2+2xy+y^2-1) (6)x3+5x2+9x+6=(x+2)(x^2+3x+3) 利用餘式定理
§解方程式(8) 方程式x2-5x+2√(x2-5x+3)=12的實根為
x^2-5x+3>0
==>x>(5+√13)/2或x<(5-√13)/2令y=x^2-5x
原式改為y+2√(y+3)=12
==>2√(y+3)=(12-y)
==>4(y+3)=(y-12)^2
==>(y-22)(y-6)=0
==>y=22或y=6
x^2-5x=22或x^2-5x=6
==>x=(5+-√113)/2或x=6或x=-1


2012-08-03 20:32:29 補充：
修正4.答案
=(x-y)(x^2+2xy+y^2-1)
=(x-y)[(x+y)^2-1]
=(x-y)(x+y+1)(x+y-1)

2012-08-03 21:12:51 補充：
(1)(xy-1)²＋(x+y)(xy+1)+5xy
=(xy+1)^2+(x+y)(xy+1)+xy
=(xy+1+x)(xy+1+y) 十字交乘
=(xy+x+1)(xy+y+1)

(5)若(x^²-y^²)^²-8(x^2+y^2-2)
=(x^2+y^2)^2-4x^2y^2-8(x^2+y^2)+16
=(x^2+y^2-4)^2-(2xy)^2
=(x^2+y^2-4+2xy)(x^2+y^2-4-2xy)
=[(x+y)^2-2^2][(x-y)^2-2^2]
=(x+y+2)(x+y-2)(x-y+2)(x-y-2)

2012-08-03 21:13:02 補充：
§解方程式
(7) 解方程式4(x²)²-13x²+1=0
==>4(x^2)^2-4x^2+1-9x^2=0
==>(2x^2-1)^2-(3x)^2=0
==>(2x^2+3x-1)(2x^2-3x-1)=0
==>x=(-3+-√17)/4 ,x=(3+-√17)/4

(9) 若解方程式(√3-1)x²-4x-(3+√3)=0
==>[(√3-1)x-(3+√3)](x+1)=0
==>x=3+2√3 ,x=-1

2012-08-03 21:27:09 補充：
更正9.
==>2x^2-4(√3+1)x+(3+√3)](√3+1)=0
==>x^2-2(√3+1)x+(3+2√3)=0
==>(x-(2+√3))(x-√3)=0
==>x=2+√3 或x=√3

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