204 Вправа 204Алгебра 8 клас Тарасенкова 2025 · § 5 · стор. 59

204. Спростіть вираз:

1) Дія в дужках:

${\displaystyle \frac{2x\left(1+3y\right)+2x\left(1-3y\right)}{(1-3y)(1+3y)}=}$${\displaystyle \frac{2x+6xy+2x-6xy}{(1-3y)(1+3y)}=}$${\displaystyle \frac{4x}{(1-3y)(1+3y)}}$;

${\displaystyle 9y^2-6y+1=}$${\displaystyle {(1-3y)}^2}$;

${\displaystyle \frac{4x}{(1-3y)(1+3y)}\cdot \frac{{(1-3y)}^2}{4x^2}=}$${\displaystyle \frac{\cancel{4x}\cancel{\left(1-3y\right)}(1-3y)}{\cancel{\left(1-3y\right)}(1+3y)\cdot \cancel{4x}x}=}$${\displaystyle \frac{1-3y}{x\left(3y+1\right)}}$;

2) Перший добуток:

${\displaystyle \frac{-(x-y)(x+y)(m+n)}{(m-n)(m+n)(x-y)}=}$${\displaystyle -\frac{x+y}{m-n}}$;

дія в дужках (${\displaystyle -\frac{x}{n-m}=}$${\displaystyle \frac{x}{m-n}}$):

${\displaystyle -\frac{x+y}{m-n}+\frac{x}{m-n}=}$${\displaystyle \frac{-x-y+x}{m-n}=}$${\displaystyle -\frac{y}{m-n}}$;

${\displaystyle -\frac{y}{m-n}\cdot \frac{m-n}{2y}=}$${\displaystyle -\frac{\cancel{y}\cancel{\left(m-n\right)}}{\cancel{\left(m-n\right)}\cdot 2\cancel{y}}=}$${\displaystyle -\frac{1}{2}=}$${\displaystyle -0,5}$.