168 Вправа 168Алгебра 8 клас Бевз 2025 · § 4 · стор. 37

168. Подай у вигляді дробу вираз.

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

б) ${\displaystyle m+\frac{3cm-m^2}{m-c}=}$${\displaystyle \frac{m^{\mathrm{\setminus }c-m}}{1}-\frac{3cm-m^2}{c-m}}$=

=${\displaystyle \frac{m(c-m)-(3cm-m^2)}{c-m}=}$${\displaystyle \frac{cm-m^2-3cm+m^2}{c-m}=}$${\displaystyle \frac{-2mc}{c-m}}$;

в) ${\displaystyle 5-\frac{3x^2+5a}{a-x^2}=}$${\displaystyle \frac{5^{\mathrm{\setminus }{x}^2-a}}{1}+\frac{3x^2+5a}{x^2-a}}$=

=${\displaystyle \frac{5(x^2-a)+3x^2+5a}{x^2-a}=}$${\displaystyle \frac{5x^2-5a+3x^2+5a}{x^2-a}=}$${\displaystyle \frac{8x^2}{x^2-a}}$;

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

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

д) ${\displaystyle \frac{1^{\mathrm{\setminus }x-1}}{1}+\frac{x^2+x+2}{x-1}=}$${\displaystyle \frac{x-1+x^2+x+2}{x-1}}$=

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