Упр.914 ГДЗ Макарычев Миндюк 9 класс (Алгебра)

Рассмотрим вариант решения задания из учебника Макарычев, Миндюк, Нешков 9 класс, Просвещение: 914. Преобразуйте выражение:

1. а)
   $$\frac12+\left(\frac{3m}{1-3m}+\frac{2m}{3m+1}\right)\cdot\frac{9m^2-6m+1}{6m^2+10m}$$
   $$=\frac12+\frac{3m(3m+1)+2m(1-3m)}{(1-3m)(3m+1)}\cdot\frac{(3m-1)^2}{2m(3m+5)}$$
   $$=\frac12+\frac{9m^2+3m+2m-6m^2}{-(3m+1)}\cdot\frac{3m-1}{2m(3m+5)}$$
   $$=\frac12+\frac{m(3m+5)}{-(3m+1)}\cdot\frac{3m-1}{3m(3m+5)}$$
   $$=\frac12+\frac{1-3m}{2(3m+1)}=\frac{3m+1+1-3m}{2(3m+1)}=\frac1{3m+1}.$$

2. б)
   $$\left(\frac1{x+y}-\frac{y^2}{xy^2-x^3}\right):\left(\frac{x-y}{x^2+xy}-\frac{x}{y^2+xy}\right)-\frac{x}{x+y}$$
   $$=\left(\frac1{x+y}-\frac{y^2}{x(y^2-x^2)}\right):\left(\frac{x-y}{x(x+y)}-\frac{x}{y(x+y)}\right)-\frac{x}{x+y}$$
   $$=\frac{xy-x^2-y^2}{x(x+y)(y-x)}:\frac{xy-y^2-x^2}{xy(x+y)}-\frac{x}{x+y}$$
   $$=\frac{xy-x^2-y^2}{x(x+y)(y-x)}\cdot\frac{xy(x+y)}{xy-y^2-x^2}-\frac{x}{x+y}$$
   $$=\frac{y}{y-x}-\frac{x}{x+y}=\frac{y(x+y)-x(y-x)}{(y-x)(x+y)}$$
   $$=\frac{xy+y^2-xy+x^2}{y^2-x^2}=\frac{x^2+y^2}{y^2-x^2}.$$

3. в)
   $$\frac{2a+3}{2a-3}\cdot\left(\frac{2a^2+3a}{4a^2+12a+9}-\frac{3a+2}{2a+3}\right)+\frac{4a-1}{2a-3}-\frac{a-1}{a}$$
   $$=\frac{2a+3}{2a-3}\cdot\frac{2a^2+3a-(3a+2)(2a+3)}{(2a+3)^2}+\frac{4a-1}{2a-3}-\frac{a-1}{a}$$
   $$=\frac{2a^2+3a-6a^2-4a-9a-6}{(2a-3)(2a+3)}+\frac{4a-1}{2a-3}-\frac{a-1}{a}$$
   $$=\frac{-4a^2-10a-6+(4a-1)(2a+3)}{(2a-3)(2a+3)}-\frac{a-1}{a}$$
   $$=\frac{-4a^2-10a-6+8a^2+12a-2a-3}{(2a-3)(2a+3)}-\frac{a-1}{a}$$
   $$=\frac{4a^2-9}{4a^2-9}-\frac{a-1}{a}=1-\frac{a-1}{a}=\frac1a.$$

4. г)
   $$\left(\frac{a+3}{a^2+2a+1}+\frac{a-1}{a^2-2a-3}\right)\cdot\frac{a^2-2a-3}{a+2}-1$$
   $$=\left(\frac{a+3}{(a+1)^2}+\frac{a-1}{(a+1)(a-3)}\right)\cdot\frac{(a+1)(a-3)}{a+2}-1$$
   $$=\frac{(a+3)(a-3)+(a-1)(a+1)}{a+1}\cdot\frac1{a+2}-1$$
   $$=\frac{a^2-9+a^2-1}{(a+1)(a+2)}-1=\frac{2a^2-10}{a^2+3a+2}-1$$
   $$=\frac{2a^2-10-a^2-3a-2}{a^2+3a+2}=\frac{a^2-3a-12}{a^2+3a+2}.$$

5. д)
   $$\frac{3(m+3)}{m^2+3m+9}+\frac{m^3-3m^2}{(m+3)^2}\cdot\left(\frac{3m}{m^3-27}+\frac1{m-3}\right)$$
   $$=\frac{3(m+3)}{m^2+3m+9}+\frac{m^2(m-3)}{(m+3)^2}\cdot\left(\frac{3m}{(m-3)(m^2+3m+9)}+\frac1{m-3}\right)$$
   $$=\frac{3(m+3)}{m^2+3m+9}+\frac{m^2(m-3)}{(m+3)^2}\cdot\frac{3m+m^2+3m+9}{(m-3)(m^2+3m+9)}$$
   $$=\frac{3(m+3)}{m^2+3m+9}+\frac{m^2}{(m+3)^2}\cdot\frac{m^2+6m+9}{m^2+3m+9}$$
   $$=\frac{3(m+3)}{m^2+3m+9}+\frac{m^2}{m^2+3m+9}=\frac{3m+9+m^2}{m^2+3m+9}=1.$$

6. е)
   $$\left(\frac{9x^2+8}{27x^3-1}-\frac1{3x-1}+\frac4{9x^2+3x+1}\right)\cdot\frac{3x-1}{3x+1}$$
   $$=\left(\frac{9x^2+8}{(3x-1)(9x^2+3x+1)}-\frac1{3x-1}+\frac4{9x^2+3x+1}\right)\cdot\frac{3x-1}{3x+1}$$
   $$=\frac{9x^2+8-(9x^2+3x+1)+4(3x-1)}{(3x-1)(9x^2+3x+1)}\cdot\frac{3x-1}{3x+1}$$
   $$=\frac{9x+3}{(3x-1)(9x^2+3x+1)}\cdot\frac{3x-1}{3x+1}=\frac{3}{9x^2+3x+1}.$$

Ответ

а) $$\frac1{3m+1}$$; б) $$\frac{x^2+y^2}{y^2-x^2}$$; в) $$\frac1a$$; г) $$\frac{a^2-3a-12}{a^2+3a+2}$$; д) $$1$$; е) $$\frac3{9x^2+3x+1}$$.