353 Вправа 353Алгебра 8 клас Бевз 2025 · § 7 · стор. 70

353. Знайди значення виразу.

а) ${\displaystyle (\frac{xy}{x^2-y^2}+\frac{y}{2y-2x})\cdot (\frac{x^2}{2y}-\frac{y}{2})}$, якщо x = 5,6, y = −2,4;

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

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

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

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

3) ${\displaystyle \frac{y}{2(x+y)}\cdot \frac{(x-y)(x+y)}{2y}=}$${\displaystyle \frac{y\cdot (x-y)(x+y)}{2(x+y)\cdot 2y}}$=

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

якщо x = 5,6, y = −2,4, то ${\displaystyle \frac{5,6-(-2,4)}{4}}$=

=${\displaystyle \frac{8}{4}=}$${\displaystyle 2}$;

б) ${\displaystyle (\frac{1}{y^2}-\frac{1}{x^2})\cdot (\frac{x+y}{x-y}-\frac{x-y}{x+y})}$, якщо x = 1/3, y = 12;

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

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

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

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

=${\displaystyle \frac{4\cancel{y}\cancel{x}\cancel{\left(y+x\right)}\cancel{\left(x-y\right)}}{\cancel{y}y\cancel{x}x\cancel{\left(y+x\right)}\cancel{\left(x-y\right)}}=}$${\displaystyle \frac{4}{yx}}$;

якщо x = 1/3, y = 12, то ${\displaystyle -\frac{4}{\frac{1}{3}\cdot 12}}$=

=${\displaystyle -\frac{4}{4}=}$${\displaystyle -1}$.