Упр.27 Вариант 3 Дидактические материалы ГДЗ Мерзляк Полонский 8 класс (Алгебра)

Задача

1) (4x^2+9y^2)/(4x^2-9y^2 )-3y/(2x+3y)+3y/(3y-2x);
2) (x+6)/(5x-10)-3/x-(26-5x)/(5x^2-10x);
3) (c+1)/(2c^2-24c+72)-1/(7c-42);
4) (y+3)/(2y+2)-(y+1)/(2y-2)+3/(y^2-1);
5) (a+1)/(a^2+a+1)-1/(a-1)+(a^3+a+1)/(a^3-1).

Подробный ответ

$$\frac{4x^2+9y^2}{4x^2-9y^2}-\frac{3y}{2x+3y}+\frac{3y}{3y-2x}$$
$$=\frac{4x^2+9y^2}{(2x-3y)(2x+3y)}-\frac{3y}{2x+3y}-\frac{3y}{2x-3y}$$
$$=\frac{4x^2+9y^2-3y(2x-3y)-3y(2x+3y)}{(2x-3y)(2x+3y)}$$
$$=\frac{4x^2+9y^2-6xy+9y^2-6xy-9y^2}{(2x-3y)(2x+3y)}$$
$$=\frac{4x^2-12xy+9y^2}{(2x-3y)(2x+3y)}$$
$$=\frac{(2x-3y)^2}{(2x-3y)(2x+3y)}=\frac{2x-3y}{2x+3y}.$$
$$\frac{x+6}{5x-10}-\frac{3}{x}-\frac{26-5x}{5x^2-10x}$$
$$=\frac{x+6}{5(x-2)}-\frac{3}{x}-\frac{26-5x}{5x(x-2)}$$
$$=\frac{x(x+6)-3\cdot 5(x-2)-(26-5x)}{5x(x-2)}$$
$$=\frac{x^2+6x-15x+30-26+5x}{5x(x-2)}$$
$$=\frac{x^2-4x+4}{5x(x-2)}$$
$$=\frac{(x-2)^2}{5x(x-2)}=\frac{x-2}{5x}.$$
$$\frac{c+1}{2c^2-24c+72}-\frac{1}{7c-42}$$
$$=\frac{c+1}{2(c-6)^2}-\frac{1}{7(c-6)}$$
$$=\frac{7(c+1)-2(c-6)}{14(c-6)^2}$$
$$=\frac{7c+7-2c+12}{14(c-6)^2}$$
$$=\frac{5c+19}{14(c-6)^2}.$$
$$\frac{y+3}{2y+2}-\frac{y+1}{2y-2}+\frac{3}{y^2-1}$$
$$=\frac{y+3}{2(y+1)}-\frac{y+1}{2(y-1)}+\frac{3}{(y-1)(y+1)}$$
$$=\frac{(y+3)(y-1)-(y+1)^2+6}{2(y-1)(y+1)}$$
$$=\frac{y^2+2y-3-y^2-2y-1+6}{2(y-1)(y+1)}$$
$$=\frac{2}{2(y-1)(y+1)}=\frac{1}{y^2-1}.$$
$$\frac{a+1}{a^2+a+1}-\frac{1}{a-1}+\frac{a^3+a+1}{a^3-1}$$
$$=\frac{a+1}{a^2+a+1}-\frac{1}{a-1}+\frac{a^3+a+1}{(a-1)(a^2+a+1)}$$
$$=\frac{(a+1)(a-1)-(a^2+a+1)+a^3+a+1}{(a-1)(a^2+a+1)}$$
$$=\frac{a^2-1-a^2-a-1+a^3+a+1}{a^3-1}$$
$$=\frac{a^3-1}{a^3-1}=1.$$