166 Вправа 166Алгебра 8 клас Тарасенкова 2025 · § 4 · стор. 50

166. Подайте у вигляді раціонального дробу.

1) ${\displaystyle x+\frac{1}{x}=}$${\displaystyle \frac{x\cdot x+1}{x}=}$${\displaystyle \frac{x^2+1}{x}}$;

2) ${\displaystyle 3x+\frac{3}{x}=}$${\displaystyle \frac{3x\cdot x+3}{x}=}$${\displaystyle \frac{3x^2+3}{x}}$;

3) ${\displaystyle \frac{1}{x^2}+4x=}$${\displaystyle \frac{1+4x\cdot x^2}{x^2}=}$${\displaystyle \frac{1+4x^3}{x^2}}$;

4) ${\displaystyle \frac{7}{y}-y^2=}$${\displaystyle \frac{7-y^2\cdot y}{y}=}$${\displaystyle \frac{7-y^3}{y}}$;

5) ${\displaystyle \frac{4-2a}{a}-2a=}$${\displaystyle \frac{4-2a-2a\cdot a}{a}=}$${\displaystyle \frac{4-2a-2a^2}{a}}$;

6) ${\displaystyle \frac{b}{7-b}-9b=}$${\displaystyle \frac{b-9b\cdot \left(7-b\right)}{7-b}=}$${\displaystyle \frac{b-63b+9b^2}{7-b}=}$${\displaystyle \frac{9b^2-62b}{7-b}}$;

7) ${\displaystyle 5a-\frac{10a^2}{2a-1}=}$${\displaystyle \frac{5a\cdot \left(2a-1\right)-10a^2}{2a-1}=}$${\displaystyle \frac{10a^2-5a-10a^2}{2a-1}=}$${\displaystyle \frac{-5a}{2a-1}}$;

8) ${\displaystyle \frac{2c}{c-5}-2c-1=}$${\displaystyle \frac{2c-(2c+1)\cdot \left(c-5\right)}{c-5}=}$${\displaystyle \frac{2c-(2c^2-9c-5)}{c-5}=}$${\displaystyle \frac{-2c^2+11c+5}{c-5}}$;

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