高専の数学解説

第1章 1.40

(1)

2a+ba2−b2+3a−b−2a+b=2a+b+3(a+b)−2(a−b)a2−b2=3a+6ba2−b2\frac{2a+b}{a^2-b^2} + \frac{3}{a-b} - \frac{2}{a+b} = \frac{2a+b+3(a+b)-2(a-b)}{a^2-b^2} = \frac{3a+6b}{a^2-b^2}

(2)

x−yxy+y−zyz+z−xzx=(x−y)z+x(y−z)+y(z−x)xyz=0\frac{x-y}{xy} + \frac{y-z}{yz} + \frac{z-x}{zx} = \frac{(x-y)z + x(y-z) + y(z-x)}{xyz} = 0

(3)

a(a−b)(a−c)+b(b−c)(b−a)+c(c−a)(c−b)=a(a−b)(a−c)+−b(b−c)(a−b)+c(a−c)(b−c)\frac{a}{(a-b)(a-c)} + \frac{b}{(b-c)(b-a)} + \frac{c}{(c-a)(c-b)} = \frac{a}{(a-b)(a-c)} + \frac{-b}{(b-c)(a-b)} + \frac{c}{(a-c)(b-c)}

=a(b−c)−b(a−c)+c(a−b)(a−b)(b−c)(a−c)=0= \frac{a(b-c) - b(a-c) + c(a-b)}{(a-b)(b-c)(a-c)} = 0

(4)

a2(a−b)(a−c)+b2(b−c)(b−a)+c2(c−a)(c−b)=a2(a−b)(a−c)+−b2(b−c)(a−b)+c2(a−c)(b−c)\frac{a^2}{(a-b)(a-c)} + \frac{b^2}{(b-c)(b-a)} + \frac{c^2}{(c-a)(c-b)} = \frac{a^2}{(a-b)(a-c)} + \frac{-b^2}{(b-c)(a-b)} + \frac{c^2}{(a-c)(b-c)}

=a2(b−c)−b2(a−c)+c2(a−b)(a−b)(b−c)(a−c)=(a−b)(b−c)(a−c)(a−b)(b−c)(a−c)=1= \frac{a^2(b-c) - b^2(a-c) + c^2(a-b)}{(a-b)(b-c)(a-c)} = \frac{(a-b)(b-c)(a-c)}{(a-b)(b-c)(a-c)} = 1

(5)

1a−b+1a+b+2aa2+b2+4a3a4+b4=2aa2−b2+2aa2+b2+4a3a4+b4=4a3a4−b4+4a3a4+b4=8a7a8−b8\frac{1}{a-b} + \frac{1}{a+b} + \frac{2a}{a^2+b^2} + \frac{4a^3}{a^4+b^4} = \frac{2a}{a^2-b^2} + \frac{2a}{a^2+b^2} + \frac{4a^3}{a^4+b^4} = \frac{4a^3}{a^4-b^4} + \frac{4a^3}{a^4+b^4} = \frac{8a^7}{a^8-b^8}

(6)

1a2−ab−ca+bc+1b2−bc−ab+ca+1c2−ca−bc+ab=1(a−b)(a−c)+−1(b−c)(a−b)+1(a−c)(b−c)\frac{1}{a^2-ab-ca+bc} + \frac{1}{b^2-bc-ab+ca} + \frac{1}{c^2-ca-bc+ab} = \frac{1}{(a-b)(a-c)} + \frac{-1}{(b-c)(a-b)} + \frac{1}{(a-c)(b-c)}

=(b−c)−(a−c)+(a−b)(a−b)(a−c)(b−c)=0= \frac{(b-c) - (a-c) + (a-b)}{(a-b)(a-c)(b-c)} = 0

解説: r31bn1z