Упр.14.14 ГДЗ Мордкович Семенов 8 класс (Алгебра)

а) 1/(b — 5)^2 — 2/(b^2 — 25) + 1/(b + 5)^2;
б) 1/(2m — 5n)^2 — 2/(25n^2 — 4m^2) + 1/(5n + 2m)^2.

а) Приведём дроби к общему знаменателю:

$$ \frac{1}{(b-5)^2}-\frac{2}{b^2-25}+\frac{1}{(b+5)^2} = \frac{1}{(b-5)^2}-\frac{2}{(b-5)(b+5)}+\frac{1}{(b+5)^2} $$

$$ = \frac{(b+5)^2-2(b-5)(b+5)+(b-5)^2}{(b-5)^2(b+5)^2} $$

$$ \begin{aligned} (b+5)^2-2(b-5)(b+5)+(b-5)^2 &= (b^2+10b+25)-2(b^2-25)+(b^2-10b+25) \\ &= b^2+10b+25-2b^2+50+b^2-10b+25 \\ &= 100 \end{aligned} $$

$$ \frac{100}{(b-5)^2(b+5)^2}=\frac{100}{(b^2-25)^2} $$

б) Аналогично:

$$ \frac{1}{(2m-5n)^2}-\frac{2}{25n^2-4m^2}+\frac{1}{(5n+2m)^2} = \frac{1}{(5n-2m)^2}-\frac{2}{(5n-2m)(5n+2m)}+\frac{1}{(5n+2m)^2} $$

$$ = \frac{(5n+2m)^2-2(5n-2m)(5n+2m)+(5n-2m)^2}{(5n-2m)^2(5n+2m)^2} $$

$$ \begin{aligned} (5n+2m)^2-2(5n-2m)(5n+2m)+(5n-2m)^2 &= (25n^2+20mn+4m^2)-2(25n^2-4m^2) \\ &\quad +(25n^2-20mn+4m^2) \\ &= 16m^2 \end{aligned} $$

$$ \frac{16m^2}{(5n-2m)^2(5n+2m)^2} = \frac{16m^2}{(25n^2-4m^2)^2} $$

Ответ

$$ \text{а) } \frac{100}{(b^2-25)^2}; \qquad \text{б) } \frac{16m^2}{(25n^2-4m^2)^2} $$