Ж: 1) 3/(x+y) + 2/x = 3x+2(x+y)/[x(x+y)] = (3x+2x+2y)/[x(x+y)] = (5x+2y)/(x²+xy); 2) a/(a+1) + 1/(a−1) =; 3) 4/(x−3) − 1/(x+3) =; 4) (c−1)/(c+2) − (c−2)/(c+1) =; 5) (x+b)/(3x) − x/(x+b) =; 6) (a−b)/(a+b) + (a+b)/(a−b) =; 7) (3ab−a²)/[(a−b)(a+b)] + 2a/(a−b) =; 8) 2y/(y−x) − 3x/(y+x) =

Question

Вопрос:

Accepted Answer

Данное преобразование верно: \(\frac{3}{x+y}+\frac{2}{x}=\frac{5x+2y}{x(x+y)}=\frac{5x+2y}{x^2+xy}\).
\(\frac{a}{a+1}+\frac{1}{a-1}=\frac{a(a-1)+(a+1)}{(a+1)(a-1)}=\frac{a^2+1}{a^2-1}\).
\(\frac{4}{x-3}-\frac{1}{x+3}=\frac{4(x+3)-(x-3)}{(x-3)(x+3)}=\frac{3x+15}{x^2-9}\).
\(\frac{c-1}{c+2}-\frac{c-2}{c+1}=\frac{(c-1)(c+1)-(c-2)(c+2)}{(c+2)(c+1)}=\frac{3}{(c+2)(c+1)}\).
\(\frac{x+b}{3x}-\frac{x}{x+b}=\frac{(x+b)^2-3x^2}{3x(x+b)}=\frac{-2x^2+2bx+b^2}{3x(x+b)}\).
\(\frac{a-b}{a+b}+\frac{a+b}{a-b}=\frac{(a-b)^2+(a+b)^2}{(a+b)(a-b)}=\frac{2(a^2+b^2)}{a^2-b^2}\).
\(\frac{3ab-a^2}{(a-b)(a+b)}+\frac{2a}{a-b}=\frac{3ab-a^2+2a(a+b)}{(a-b)(a+b)}=\frac{a(a+5b)}{(a-b)(a+b)}\).
\(\frac{2y}{y-x}-\frac{3x}{y+x}=\frac{2y(y+x)-3x(y-x)}{(y-x)(y+x)}=\frac{3x^2-xy+2y^2}{y^2-x^2}\).

Ответ: выражения упрощены приведением дробей к общему знаменателю.