201 Вправа 201Алгебра 8 клас Бевз 2025 · § 4 · стор. 41

201. Виконай дії.

а) ${\displaystyle \frac{1}{x^2+3xy}+\frac{2}{9y^2-x^2}+\frac{1}{2x-6y}}$=

=${\displaystyle \frac{1^{\mathrm{\setminus }2\left(x-3y\right)}}{x(x+3y)}-\frac{2^{\mathrm{\setminus }2x}}{(x-3y)(x+3y)}+\frac{1^{\mathrm{\setminus }x\left(x+3y\right)}}{2(x-3y)}}$=

=${\displaystyle \frac{2(x-3y)-4x+x(x+3y)}{2x(x-3y)(x+3y)}=}$${\displaystyle \frac{2x-6y-4x+x^2+3xy}{2x(x-3y)(x+3y)}}$=

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

б) ${\displaystyle \frac{7-x^2}{x^3-1}+\frac{3x+5}{x^2+x+1}+\frac{2}{1-x}}$=

=${\displaystyle \frac{7-x^2}{(x-1)(x^2+x+1)}+\frac{{3x+5}^{\mathrm{\setminus }x-1}}{x^2+x+1}-\frac{2^{\mathrm{\setminus }{x}^2+x+1}}{x-1}}$=

=${\displaystyle \frac{7-x^2+(3x+5)(x-1)-2(x^2+x+1)}{(x-1)(x^2+x+1)}}$=

=${\displaystyle \frac{7-x^2+3x^2+2x-5-2x^2-2x-2}{(x-1)(x^2+x+1)}}$=

=${\displaystyle \frac{0}{(x-1)(x^2+x+1)}=}$${\displaystyle 0}$.