322 Вправа 322Алгебра 8 клас Бевз 2025 · § 7 · стор. 67

322. Установіть відповідність між виразами (1–3) та значеннями цих виразів (А–Д) за умови, що x = −2, y = 2.

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

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

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

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

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

якщо x = −2, y = 2, то ${\displaystyle \frac{-2}{-2-2}=}$${\displaystyle 0,5}$ — В;

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

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

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

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

=${\displaystyle \frac{-8xy}{(2x-y)(2x+y)}=}$${\displaystyle \frac{8xy}{(2x+y)(y-2x)}}$;

${\displaystyle \frac{8xy}{(2x+y)(y-2x)}:\frac{4xy}{2x-y}=}$${\displaystyle \frac{8xy}{(2x+y)(y-2x)}\cdot \frac{2x-y}{4xy}}$=

=${\displaystyle \frac{8xy\cdot (2x-y)}{(2x+y)(y-2x)\cdot 4xy}}$=

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

якщо x = −2, y = 2, то ${\displaystyle -\frac{2}{2\cdot (-2)+2}}$=

=${\displaystyle -\frac{2}{-2}=}$${\displaystyle 1}$ — Г;

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

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

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

=${\displaystyle \frac{x^2+4xy+4y^2+x^2-4xy+4y^2}{(x-2y)(x+2y)}=}$${\displaystyle \frac{2x^2+8y^2}{(x-2y)(x+2y)}}$=

=${\displaystyle \frac{2(x^2+4y^2)}{(x-2y)(x+2y)}=}$${\displaystyle \frac{2x^2+8y^2}{x^2-4y^2}}$;

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

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

=${\displaystyle \frac{6x\cancel{\left(x+2y\right)}\cancel{\left(x^2+4y^2\right)}}{\cancel{\left(x+2y\right)}\cancel{\left(x^2+4y^2\right)}(x-2y)}=}$${\displaystyle \frac{6x}{x-2y}}$;

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

=${\displaystyle \frac{-12}{-6}=}$${\displaystyle 2}$ — Д.

Відповідь: 1 — В, 2 — Г, 3 — Д.