127 Вправа 127Алгебра 8 клас Тарасенкова 2025 · § 3 · стор. 40

127. Зведіть до спільного знаменника дроби.

Кожен знаменник розкладаємо на:

1) ${\displaystyle \frac{1}{a+b}}$, ${\displaystyle \frac{1}{a^2-ab}}$, ${\displaystyle \frac{1}{a^2-b^2}}$;

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

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

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

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

2) ${\displaystyle \frac{1}{b^2-4}}$, ${\displaystyle \frac{1}{3b+6}}$, ${\displaystyle \frac{1}{2-b}}$;

${\displaystyle b^2-4=}$${\displaystyle (b-2)(b+2)}$; ${\displaystyle 3b+6=}$${\displaystyle 3(b+2)}$; ${\displaystyle 2-b=}$${\displaystyle -(b-2)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }3}}{(b-2)(b+2)}=}$${\displaystyle \frac{1\cdot 3}{(b-2)(b+2)\cdot 3}=}$${\displaystyle \frac{3}{3\left(b^2-4\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }b-2}}{3(b+2)}=}$${\displaystyle \frac{1\cdot \left(b-2\right)}{3(b+2)\cdot \left(b-2\right)}=}$${\displaystyle \frac{b-2}{3\left(b^2-4\right)}}$;

${\displaystyle \frac{1}{2-b}=}$${\displaystyle \frac{{-1}^{\mathrm{\setminus }3\left(b+2\right)}}{b-2}=}$${\displaystyle \frac{-1\cdot 3\left(b+2\right)}{(b-2)\cdot 3\left(b+2\right)}=}$${\displaystyle \frac{-3b-6}{3\left(b^2-4\right)}}$;

3) ${\displaystyle \frac{1}{2y-3}}$, ${\displaystyle \frac{1}{9-4y^2}}$, ${\displaystyle \frac{1}{4y+6}}$;

${\displaystyle 9-4y^2=}$${\displaystyle -(2y-3)(2y+3)}$; ${\displaystyle 4y+6=}$${\displaystyle 2(2y+3)}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2\left(2y+3\right)}}{2y-3}=}$${\displaystyle \frac{1\cdot 2\left(2y+3\right)}{(2y-3)\cdot 2\left(2y+3\right)}=}$${\displaystyle \frac{4y+6}{2\left(4y^2-9\right)}}$;

${\displaystyle \frac{1}{9-4y^2}=}$${\displaystyle \frac{{-1}^{\mathrm{\setminus }2}}{(2y-3)(2y+3)}=}$${\displaystyle \frac{-1\cdot 2}{(2y-3)(2y+3)\cdot 2}=}$${\displaystyle \frac{-2}{2\left(4y^2-9\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }2y-3}}{2(2y+3)}=}$${\displaystyle \frac{1\cdot \left(2y-3\right)}{2(2y+3)\cdot \left(2y-3\right)}=}$${\displaystyle \frac{2y-3}{2\left(4y^2-9\right)}}$;

4) ${\displaystyle \frac{1}{3x-1}}$, ${\displaystyle \frac{1}{9x^2-1}}$, ${\displaystyle \frac{1}{9x^2-6x+1}}$;

${\displaystyle 9x^2-1=}$${\displaystyle (3x-1)(3x+1)}$; ${\displaystyle 9x^2-6x+1=}$${\displaystyle {(3x-1)}^2}$;

${\displaystyle \frac{1^{\mathrm{\setminus }\left(3x-1\right)\left(3x+1\right)}}{3x-1}=}$${\displaystyle \frac{1\cdot \left(3x-1\right)\left(3x+1\right)}{(3x-1)\cdot \left(3x-1\right)\left(3x+1\right)}=}$${\displaystyle \frac{9x^2-1}{{\left(3x-1\right)}^2\left(3x+1\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }3x-1}}{(3x-1)(3x+1)}=}$${\displaystyle \frac{1\cdot \left(3x-1\right)}{(3x-1)(3x+1)\cdot \left(3x-1\right)}=}$${\displaystyle \frac{3x-1}{{\left(3x-1\right)}^2\left(3x+1\right)}}$;

${\displaystyle \frac{1^{\mathrm{\setminus }3x+1}}{{(3x-1)}^2}=}$${\displaystyle \frac{1\cdot \left(3x+1\right)}{{(3x-1)}^2\cdot \left(3x+1\right)}=}$${\displaystyle \frac{3x+1}{{\left(3x-1\right)}^2\left(3x+1\right)}}$.