192 Вправа 192Алгебра 8 клас Бевз 2025 · § 4 · стор. 40

192. Знайди значення виразу.

а) ${\displaystyle \frac{2a}{1-a^2}+\frac{a+1}{2a-2}-\frac{a-1}{3a+3}}$=

=${\displaystyle -\frac{{2a}^{\mathrm{\setminus }6}}{(a-1)(a+1)}+\frac{{a+1}^{\mathrm{\setminus }3\left(a+1\right)}}{2(a-1)}-\frac{{a-1}^{\mathrm{\setminus }2\left(a-1\right)}}{3(a+1)}}$=

=${\displaystyle \frac{-12a+3(a+1)(a+1)-2(a-1)(a-1)}{6(a-1)(a+1)}}$=

=${\displaystyle \frac{-12a+3a^2+6a+3-2a^2+4a-2}{6(a-1)(a+1)}=}$${\displaystyle \frac{a^2-2a+1}{6(a-1)(a+1)}}$=

=${\displaystyle \frac{(a-1)\cancel{\left(a-1\right)}}{6(a+1)\cancel{\left(a-1\right)}}=}$${\displaystyle \frac{a-1}{6(a+1)}}$;

якщо a = 3, то ${\displaystyle \frac{3-1}{6(3+1)}=}$${\displaystyle \frac{1}{12}}$;

б) ${\displaystyle \frac{x+2}{x}-\frac{x}{x-2}+\frac{x+2}{x^2-2x}}$=

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

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

якщо x = 1/5, то ${\displaystyle \frac{1}{(0,2)}=}$${\displaystyle 5}$;

в) ${\displaystyle \frac{1}{x}+\frac{x+2y}{x^2-2xy}-\frac{4x}{x^2-4y^2}}$=

=${\displaystyle \frac{1^{\mathrm{\setminus }\left(x-2y\right)\left(x+2y\right)}}{x}+\frac{{x+2y}^{\mathrm{\setminus }x+2y}}{x(x-2y)}-\frac{{4x}^{\mathrm{\setminus }x}}{(x-2y)(x+2y)}}$=

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

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

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

якщо x = 2, y = 3, то ${\displaystyle -\frac{2}{2+2\cdot 3}=}$${\displaystyle -0,25}$.