309 Вправа 309Алгебра 8 клас Бевз 2025 · § 7 · стор. 65

309. Спрости вираз.

а) ${\displaystyle \frac{a-b}{a}\cdot (\frac{a}{a-b}+\frac{a}{b})}$;

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

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

2) ${\displaystyle \frac{a-b}{a}\cdot \frac{a^2}{b(a-b)}=}$${\displaystyle \frac{(a-b)\cdot a^2}{a\cdot b(a-b)}=}$${\displaystyle \frac{\cancel{a}a\cancel{\left(a-b\right)}}{\cancel{a}b\cancel{\left(a-b\right)}}=}$${\displaystyle \frac{a}{b}}$;

б) ${\displaystyle (\frac{x}{y}-\frac{x}{x+y})\cdot \frac{x+y}{x}}$;

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

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

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

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

1) ${\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{2x^2+2y^2}{(x-y)(x+y)}}$=

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

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

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

г) ${\displaystyle (\frac{a-b}{a+b}-\frac{a+b}{a-b}):\frac{4}{a^2-b^2}}$;

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

=${\displaystyle \frac{a^2-2ab+b^2-a^2-2ab-b^2}{(a-b)(a+b)}}$=

=${\displaystyle \frac{-4ab}{(a-b)(a+b)}=}$${\displaystyle \frac{4ab}{(a+b)(b-a)}}$;

2) ${\displaystyle \frac{4ab}{(a+b)(b-a)}:\frac{4}{a^2-b^2}=}$${\displaystyle \frac{4ab}{(a+b)(b-a)}\cdot \frac{a^2-b^2}{4}}$=

=${\displaystyle \frac{4ab\cdot (a^2-b^2)}{(a+b)(b-a)\cdot 4}=}$${\displaystyle -\frac{\cancel{4}ab\cancel{\left(a+b\right)}\cancel{\left(b-a\right)}}{\cancel{4}\cancel{\left(a+b\right)}\cancel{\left(b-a\right)}}=}$${\displaystyle -ab}$.