337 Вправа 337Алгебра 8 клас Бевз 2025 · § 7 · стор. 69

337. Спрости вираз.

а) ${\displaystyle \frac{n^2-1}{n^2+n}\cdot {(1-\frac{1}{1-\frac{1}{n}})}^2}$;

1) ${\displaystyle \frac{1^{\mathrm{\setminus }n}}{1}-\frac{1}{n}=}$${\displaystyle \frac{n-1}{n}}$;

2) ${\displaystyle 1:\frac{n-1}{n}=}$${\displaystyle 1\cdot \frac{n}{n-1}=}$${\displaystyle \frac{n}{n-1}}$;

3) ${\displaystyle \frac{1^{\mathrm{\setminus }n-1}}{1}-\frac{n}{n-1}=}$${\displaystyle \frac{n-1-n}{n-1}=}$${\displaystyle \frac{-1}{n-1}=}$${\displaystyle \frac{1}{1-n}}$;

4) ${\displaystyle {\left(\frac{1}{1-n}\right)}^2=}$${\displaystyle \frac{1^2}{{(1-n)}^2}=}$${\displaystyle \frac{1}{{(1-n)}^2}}$;

5) ${\displaystyle \frac{n^2-1}{n^2+n}\cdot \frac{1}{{(n-1)}^2}=}$${\displaystyle \frac{n^2-1}{(n^2+n)\cdot {(n-1)}^2}}$=

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

б) ${\displaystyle \frac{a^4+c^4}{{(a-c)}^2}-\frac{2}{{(\frac{1}{a}-\frac{1}{c})}^2}}$;

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

2) ${\displaystyle {\left(\frac{c-a}{ac}\right)}^2=}$${\displaystyle \frac{{(c-a)}^2}{{\left(ac\right)}^2}=}$${\displaystyle \frac{{(c-a)}^2}{c^2{a}^2}}$;

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

4) ${\displaystyle \frac{a^4+c^4}{{(a-c)}^2}-\frac{2a^2{c}^2}{{(a-c)}^2}=}$${\displaystyle \frac{a^4+c^4-2a^2{c}^2}{{(a-c)}^2}}$=

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