高専の数学解説

第1章 1.38

(1)

x2−(y−z)2(x−y)2−z2={x−(y−z)}(x+(y−z)){(x−y)−z}{(x−y)+z}=x+y−zx−y−z\frac{x^2-(y-z)^2}{(x-y)^2-z^2}=\frac{\{x-(y-z)\}(x+(y-z))}{\{(x-y)-z\}\{(x-y)+z\}}=\frac{x+y-z}{x-y-z}

(2)

a3−a2−8a+12a2−4a+4=(a−2)2(a+3)(a−2)2=a+3\frac{a^3-a^2-8a+12}{a^2-4a+4}=\frac{(a-2)^2(a+3)}{(a-2)^2}=a+3

(3)

2x3+x2+x−1x3−1=(2x−1)(x2+x+1)(x−1)(x2+x+1)=2x−1x−1\frac{2x^3+x^2+x-1}{x^3-1}=\frac{(2x-1)(x^2+x+1)}{(x-1)(x^2+x+1)}=\frac{2x-1}{x-1}

(4)

x3+3x2−x−3x4+x3+x2−x−2=(x−1)(x+3)(x+1)(x−1)(x3+2x2+3x+2)=(x−1)(x+3)(x+1)(x−1)(x+1)(x2+x+2)=x+3x2+x+2\frac{x^3+3x^2-x-3}{x^4+x^3+x^2-x-2}=\frac{(x-1)(x+3)(x+1)}{(x-1)(x^3+2x^2+3x+2)}=\frac{(x-1)(x+3)(x+1)}{(x-1)(x+1)(x^2+x+2)}=\frac{x+3}{x^2+x+2}

解説: r31bn1z