13 Вправа 13Алгебра 8 клас Бевз 2025 · § 1 · стор. 9

13. Знайди значення дробу.

а) Якщо ${\displaystyle x=}$${\displaystyle 2}$, то ${\displaystyle \frac{x^2}{x^2-10}=}$${\displaystyle \frac{2^2}{2^2-10}=}$${\displaystyle \frac{4}{4-10}}$=

=${\displaystyle \frac{4}{-6}=}$${\displaystyle -\frac{2}{3}}$;

б) якщо ${\displaystyle x=}$${\displaystyle -12}$ і ${\displaystyle y=}$${\displaystyle 13}$, то ${\displaystyle \frac{2x}{x^2-y^2}}$=

=${\displaystyle \frac{2x}{(x-y)(x+y)}=}$${\displaystyle \frac{2\cdot (-12)}{(-12-13)(-12+13)}=}$${\displaystyle \frac{-24}{-25\cdot 1}=}$${\displaystyle \frac{24}{25}}$;

в) якщо ${\displaystyle a=}$${\displaystyle 9}$ і ${\displaystyle b=}$${\displaystyle -7}$, то ${\displaystyle \frac{2(a+b)}{a-b}}$=

=${\displaystyle \frac{2\cdot (9+(-7))}{9-(-7)}=}$${\displaystyle \frac{2\cdot 2}{9+7}=}$${\displaystyle \frac{4}{16}=}$${\displaystyle \frac{1}{4}}$;

г) якщо ${\displaystyle x=}$${\displaystyle 5}$ і ${\displaystyle y=}$${\displaystyle \frac{1}{3}}$, то ${\displaystyle \frac{x}{y(x-2)}=}$${\displaystyle \frac{5}{\frac{1}{3}\cdot (5-2)}}$=

=${\displaystyle \frac{5}{\frac{1}{3}\cdot 3}=}$${\displaystyle \frac{5}{1}=}$${\displaystyle 5}$.