119° Вправа 119Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 38

119. Зведіть дріб:

Новий знаменник записуємо як добуток, щоб побачити додатковий:

1) ${\displaystyle \frac{1}{a+b}}$:

${\displaystyle \frac{1^{\mathrm{\setminus }2}}{a+b}=}$${\displaystyle \frac{1\cdot 2}{(a+b)\cdot 2}=}$${\displaystyle \frac{2}{2\left(a+b\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }a+b}}{a+b}=}$${\displaystyle \frac{1\cdot \left(a+b\right)}{(a+b)\cdot \left(a+b\right)}=}$${\displaystyle \frac{a+b}{{\left(a+b\right)}^2}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }3a}}{a+b}=}$${\displaystyle \frac{1\cdot 3a}{(a+b)\cdot 3a}=}$${\displaystyle \frac{3a}{3a\left(a+b\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }a-b}}{a+b}=}$${\displaystyle \frac{1\cdot \left(a-b\right)}{(a+b)\cdot \left(a-b\right)}=}$${\displaystyle \frac{a-b}{a^2-b^2}}$;

2) ${\displaystyle \frac{5x}{x-y}}$:

${\displaystyle \frac{{5x}^{\mathrm{\setminus }2}}{x-y}=}$${\displaystyle \frac{5x\cdot 2}{(x-y)\cdot 2}=}$${\displaystyle \frac{10x}{2\left(x-y\right)}}$;

${\displaystyle \frac{{5x}^{\mathrm{\setminus }x-y}}{x-y}=}$${\displaystyle \frac{5x\cdot \left(x-y\right)}{(x-y)\cdot \left(x-y\right)}=}$${\displaystyle \frac{5x^2-5xy}{{\left(x-y\right)}^2}}$;

${\displaystyle \frac{{5x}^{\mathrm{\setminus }4y}}{x-y}=}$${\displaystyle \frac{5x\cdot 4y}{(x-y)\cdot 4y}=}$${\displaystyle \frac{20xy}{4y\left(x-y\right)}}$;

${\displaystyle \frac{{5x}^{\mathrm{\setminus }x+y}}{x-y}=}$${\displaystyle \frac{5x\cdot \left(x+y\right)}{(x-y)\cdot \left(x+y\right)}=}$${\displaystyle \frac{5x^2+5xy}{x^2-y^2}}$;

3) ${\displaystyle \frac{3m-2n}{2(m+n)}}$:

${\displaystyle \frac{{3m-2n}^{\mathrm{\setminus }7}}{2(m+n)}=}$${\displaystyle \frac{(3m-2n)\cdot 7}{2(m+n)\cdot 7}=}$${\displaystyle \frac{21m-14n}{14\left(m+n\right)}}$;

${\displaystyle \frac{{3m-2n}^{\mathrm{\setminus }4\left(m+n\right)}}{2(m+n)}=}$${\displaystyle \frac{(3m-2n)\cdot 4\left(m+n\right)}{2(m+n)\cdot 4\left(m+n\right)}=}$${\displaystyle \frac{(3m-2n)(4m+4n)}{8{(m+n)}^2}}$;${\displaystyle \frac{12m^2+12mn-8mn-8n^2}{8{(m+n)}^2}=}$${\displaystyle \frac{12m^2+4mn-8n^2}{8{\left(m+n\right)}^2}}$.