326 Вправа 326Алгебра 8 клас Бевз 2025 · § 7 · стор. 67

326. Спрости вираз.

а) ${\displaystyle (x+1-\frac{1}{1-x}):(x-\frac{x^2}{x-1})}$;

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

=${\displaystyle \frac{-x^2}{1-x}=}$${\displaystyle \frac{x^2}{(x-1)}}$;

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

=${\displaystyle \frac{-x}{x-1}=}$${\displaystyle \frac{x}{1-x}}$;

3) ${\displaystyle \frac{x^2}{x-1}:\frac{x}{1-x}=}$${\displaystyle \frac{x^2}{x-1}\cdot \frac{1-x}{x}=}$${\displaystyle \frac{x^2\cdot (1-x)}{(x-1)\cdot x}}$=

=${\displaystyle -\frac{\cancel{x}x\cancel{\left(x-1\right)}}{\cancel{x}\cancel{\left(x-1\right)}}=}$${\displaystyle -x}$;

б) ${\displaystyle (\frac{a}{c^2}+\frac{1}{c}+\frac{1}{a}):(\frac{a^2}{c^2}-\frac{c}{a})}$;

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

2) ${\displaystyle \frac{{a^2}^{\mathrm{\setminus }a}}{c^2}-\frac{c^{\mathrm{\setminus }{c}^2}}{a}=}$${\displaystyle \frac{a^3-c^3}{a{c}^2}=}$${\displaystyle \frac{(a-c)(a^2+ac+c^2)}{a{c}^2}}$;

3) ${\displaystyle \frac{a^2+ac+c^2}{a{c}^2}:\frac{(a-c)(a^2+ac+c^2)}{a{c}^2}}$=

=${\displaystyle \frac{a^2+ac+c^2}{a{c}^2}\cdot \frac{a{c}^2}{(a-c)(a^2+ac+c^2)}}$=

=${\displaystyle \frac{\cancel{a}\cancel{c^2}\cancel{\left(a^2+ac+c^2\right)}}{\cancel{a}\cancel{c^2}\cancel{\left(a^2+ac+c^2\right)}(a-c)}=}$${\displaystyle \frac{1}{a-c}}$.