124° Вправа 124Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 39

124. Зведіть до спільного знаменника дроби.

Розкладаємо знаменники на:

1) ${\displaystyle \frac{1}{a+1}}$ і ${\displaystyle \frac{1}{2a+2}}$;

${\displaystyle 2a+2=}$${\displaystyle 2(a+1)}$;

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

${\displaystyle \frac{1}{2a+2}}$ — без змін;

2) ${\displaystyle \frac{1}{a-2}}$ і ${\displaystyle \frac{1}{3a-6}}$;

${\displaystyle 3a-6=}$${\displaystyle 3(a-2)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }3}}{a-2}=}$${\displaystyle \frac{1\cdot 3}{(a-2)\cdot 3}=}$${\displaystyle \frac{3}{3a-6}}$;

${\displaystyle \frac{1}{3a-6}}$ — без змін;

3) ${\displaystyle \frac{1}{1-x}}$ і ${\displaystyle \frac{1}{4-4x}}$;

${\displaystyle 4-4x=}$${\displaystyle 4(1-x)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }4}}{1-x}=}$${\displaystyle \frac{1\cdot 4}{(1-x)\cdot 4}=}$${\displaystyle \frac{4}{4-4x}}$;

${\displaystyle \frac{1}{4-4x}}$ — без змін;

4) ${\displaystyle \frac{1}{2a+1}}$ і ${\displaystyle \frac{1}{4a+2}}$;

${\displaystyle 4a+2=}$${\displaystyle 2(2a+1)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2}}{2a+1}=}$${\displaystyle \frac{1\cdot 2}{(2a+1)\cdot 2}=}$${\displaystyle \frac{2}{4a+2}}$;

${\displaystyle \frac{1}{4a+2}}$ — без змін;

5) ${\displaystyle \frac{1}{2-5y}}$ і ${\displaystyle \frac{1}{4-10y}}$;

${\displaystyle 4-10y=}$${\displaystyle 2(2-5y)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2}}{2-5y}=}$${\displaystyle \frac{1\cdot 2}{(2-5y)\cdot 2}=}$${\displaystyle \frac{2}{4-10y}}$;

${\displaystyle \frac{1}{4-10y}}$ — без змін;

6) ${\displaystyle \frac{3m}{21m-14n}}$ і ${\displaystyle \frac{5n}{3m-2n}}$;

${\displaystyle 21m-14n=}$${\displaystyle 7(3m-2n)}$;

${\displaystyle \frac{3m}{21m-14n}}$ — без змін;

${\displaystyle \frac{{5n}^{\mathrm{\setminus }7}}{3m-2n}=}$${\displaystyle \frac{5n\cdot 7}{(3m-2n)\cdot 7}=}$${\displaystyle \frac{35n}{21m-14n}}$;

7) ${\displaystyle \frac{4a}{15a+3b}}$ і ${\displaystyle -\frac{6b}{5a+b}}$;

${\displaystyle 15a+3b=}$${\displaystyle 3(5a+b)}$;

${\displaystyle \frac{4a}{15a+3b}}$ — без змін;

${\displaystyle -\frac{{6b}^{\mathrm{\setminus }3}}{5a+b}=}$${\displaystyle -\frac{6b\cdot 3}{(5a+b)\cdot 3}=}$${\displaystyle -\frac{18b}{15a+3b}}$;

8) ${\displaystyle \frac{2}{16-4b}}$ і ${\displaystyle \frac{7b}{4-b}}$;

${\displaystyle 16-4b=}$${\displaystyle 4(4-b)}$;

${\displaystyle \frac{2}{16-4b}}$ — без змін;

${\displaystyle \frac{{7b}^{\mathrm{\setminus }4}}{4-b}=}$${\displaystyle \frac{7b\cdot 4}{(4-b)\cdot 4}=}$${\displaystyle \frac{28b}{16-4b}}$.