Упр.152 Повторение ГДЗ Мерзляк 11 класс Базовый уровень (Алгебра)

1) (m+1)/(m-1)-(m^2+1)/(m^2-1);
2) b^2/(2ab+a^2+b^2)+(a-b)/(a+b);
3) 3a/(9a^2-1)-(a+2)/(3a^2+a);
4) (k-2)/(k^2+6k+9)-k/(k^2-9);
5) (x-20)/(x^2+5x)+x/(x+5)-(x-5)/x;
6) (y+3)/(y-3)-(y-3)/(y+3)-36/(y^2-9).

1. $$\frac{m+1}{m-1}-\frac{m^2+1}{m^2-1}=\frac{m+1}{m-1}-\frac{m^2+1}{(m-1)(m+1)}$$
   $$=\frac{(m+1)^2-(m^2+1)}{(m-1)(m+1)}=\frac{m^2+2m+1-m^2-1}{(m-1)(m+1)}=\frac{2m}{m^2-1}$$

2. $$\frac{b^2}{2ab+a^2+b^2}+\frac{a-b}{a+b}=\frac{b^2}{(a+b)^2}+\frac{a-b}{a+b}$$
   $$=\frac{b^2+(a-b)(a+b)}{(a+b)^2}=\frac{b^2+a^2-b^2}{(a+b)^2}=\frac{a^2}{(a+b)^2}$$

3. $$\frac{3a}{9a^2-1}-\frac{a+2}{3a^2+a}=\frac{3a}{(3a-1)(3a+1)}-\frac{a+2}{a(3a+1)}$$
   $$=\frac{3a^2-(a+2)(3a-1)}{a(3a-1)(3a+1)}=\frac{3a^2-3a^2-5a+2}{a(3a-1)(3a+1)}$$
   $$=\frac{2-5a}{a(9a^2-1)}$$

4. $$\frac{k-2}{k^2+6k+9}-\frac{k}{k^2-9}=\frac{k-2}{(k+3)^2}-\frac{k}{(k-3)(k+3)}$$
   $$=\frac{(k-2)(k-3)-k(k+3)}{(k-3)(k+3)^2}=\frac{k^2-5k+6-k^2-3k}{(k-3)(k+3)^2}$$
   $$=\frac{6-8k}{(k-3)(k+3)^2}=\frac{2(3-4k)}{(k-3)(k+3)^2}$$

5. $$\frac{x-20}{x^2+5x}+\frac{x}{x+5}-\frac{x-5}{x}=\frac{x-20}{x(x+5)}+\frac{x}{x+5}-\frac{x-5}{x}$$
   $$=\frac{x-20+x^2-x^2+25}{x(x+5)}=\frac{x+5}{x(x+5)}=\frac{1}{x}$$

6. $$\frac{y+3}{y-3}-\frac{y-3}{y+3}-\frac{36}{y^2-9}=\frac{y+3}{y-3}-\frac{y-3}{y+3}-\frac{36}{(y-3)(y+3)}$$
   $$=\frac{(y+3)^2-(y-3)^2-36}{(y-3)(y+3)}=\frac{y^2+6y+9-y^2+6y-9-36}{(y-3)(y+3)}$$
   $$=\frac{12(y-3)}{(y-3)(y+3)}=\frac{12}{y+3}$$

Ответ

1) $$\frac{2m}{m^2-1}$$; 2) $$\frac{a^2}{(a+b)^2}$$; 3) $$\frac{2-5a}{a(9a^2-1)}$$; 4) $$\frac{2(3-4k)}{(k-3)(k+3)^2}$$; 5) $$\frac{1}{x}$$; 6) $$\frac{12}{y+3}$$.