第4章 4.32

(1)

x2+xy12y2+2x+15y3x^2 + xy - 12y^2 + 2x + 15y - 3

=(x+4y)(x3y)+2x+15y3= (x+4y)(x-3y) + 2x + 15y - 3

a(x+4y)+b(x3y)=2x+15y(a+b)x=2x(4a3b)y=15ya(x+4y) + b(x-3y) = 2x+15y \qquad (a+b)x = 2x \qquad (4a-3b)y = 15y

a+b=24a3b=15a+b=2 \quad ① \qquad 4a-3b=15 \quad ②

a=3,b=1a=3, b=-1

ab=3ab=-3 であるので

(x+4y1)(x3y+3)(x+4y-1)(x-3y+3)

(2)

2x25xy3y2+3x2y+12x^2 - 5xy - 3y^2 + 3x - 2y + 1

=(2x+y)(x3y)+3x2y+1= (2x+y)(x-3y) + 3x - 2y + 1

(2x+y)+(x3y)=3x2y(2x+y)+(x-3y) = 3x-2y であるので

(2x+y+1)(x3y+1)(2x+y+1)(x-3y+1)

(3)

(a21)x22(a2+1)x+a21(a^2-1)x^2 - 2(a^2+1)x + a^2-1

x=(a2+1)±(a2+1)2(a21)2a21=a2+1±4a2a21=a2+1±2aa21=(a+1)2a21,(a1)2a21=a+1a1,a1a+1x = \frac{(a^2+1) \pm \sqrt{(a^2+1)^2-(a^2-1)^2}}{a^2-1} = \frac{a^2+1\pm\sqrt{4a^2}}{a^2-1} = \frac{a^2+1\pm2a}{a^2-1} = \frac{(a+1)^2}{a^2-1}, \frac{(a-1)^2}{a^2-1} = \frac{a+1}{a-1}, \frac{a-1}{a+1}

これから

{(a1)xa1}{(a+1)xa+1}\{(a-1)x-a-1\}\{(a+1)x-a+1\}

解説: r31bn1z