316 Вправа 316Алгебра 8 клас Бевз 2025 · § 7 · стор. 66

316. Виконай дії.

а) ${\displaystyle (\frac{x+y}{y}-\frac{2y}{y-x})\cdot \frac{y-x}{x^2+y^2}}$;

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

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

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

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

б) ${\displaystyle (\frac{2a}{a+x}+\frac{x-a}{a}):\frac{x^2+a^2}{x+a}}$;

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

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

2) ${\displaystyle \frac{a^2+x^2}{a(a+x)}:\frac{x^2+a^2}{x+a}=}$${\displaystyle \frac{a^2+x^2}{a(a+x)}\cdot \frac{x+a}{x^2+a^2}}$=

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

в) ${\displaystyle (a+x-\frac{2ax}{a+x}):(\frac{a-x}{a+x}+\frac{x}{a})}$;

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

=${\displaystyle \frac{a^2+2ax+x^2-2ax}{a+x}=}$${\displaystyle \frac{a^2+x^2}{a+x}}$;

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

=${\displaystyle \frac{a^2-ax+ax+x^2}{a(a+x)}=}$${\displaystyle \frac{a^2+x^2}{a(a+x)}}$;

3) ${\displaystyle \frac{a^2+x^2}{a+x}:\frac{a^2+x^2}{a(a+x)}=}$${\displaystyle \frac{a^2+x^2}{a+x}\cdot \frac{a(a+x)}{a^2+x^2}}$=

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

г) ${\displaystyle (\frac{y}{x}-\frac{2y}{x+y})\cdot (y+\frac{x^2+y^2}{x-y})}$;

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

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

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

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

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

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