Упр.37.40 ГДЗ Мордкович 10-11 класс (Алгебра)
- а) $$\frac{a^{1/2}+b^{1/2}}{a^{1/2}}-\frac{a^{1/2}}{a^{1/2}-b^{1/2}}+\frac{b}{a-a^{1/2}b^{1/2}}$$
а)
$$\frac{a^{\frac12}+b^{\frac12}}{a^{\frac12}} -\frac{a^{\frac12}}{a^{\frac12}-b^{\frac12}} +\frac{b}{a-a^{\frac12}b^{\frac12}}$$
Приведём к общему знаменателю $$a^{\frac12}\left(a^{\frac12}-b^{\frac12}\right)\left(a-a^{\frac12}b^{\frac12}\right)$$:
$$\frac{\left(a^{\frac12}+b^{\frac12}\right)\left(a^{\frac12}-b^{\frac12}\right)\left(a-a^{\frac12}b^{\frac12}\right)}{a^{\frac12}\left(a^{\frac12}-b^{\frac12}\right)\left(a-a^{\frac12}b^{\frac12}\right)} -\frac{a^{\frac12}\cdot a^{\frac12}\left(a-a^{\frac12}b^{\frac12}\right)}{a^{\frac12}\left(a^{\frac12}-b^{\frac12}\right)\left(a-a^{\frac12}b^{\frac12}\right)} +\frac{b\cdot a^{\frac12}\left(a^{\frac12}-b^{\frac12}\right)}{a^{\frac12}\left(a^{\frac12}-b^{\frac12}\right)\left(a-a^{\frac12}b^{\frac12}\right)}$$
Так как
$$\left(a^{\frac12}+b^{\frac12}\right)\left(a^{\frac12}-b^{\frac12}\right)=a-b,$$
то числитель равен
$$(a-b)-a+b=0.$$
Следовательно,
$$0.$$
Ответ: $$0$$
б)
$$\frac{2a^{\frac13}}{a^{\frac43}-3a^{\frac13}} -\frac{a^{\frac23}}{a^{\frac53}-a^{\frac23}} -\frac{a+1}{a^2-4a+3}$$
Разложим знаменатели на множители:
$$a^{\frac43}-3a^{\frac13}=a^{\frac13}(a-3), \qquad a^{\frac53}-a^{\frac23}=a^{\frac23}(a-1), \qquad a^2-4a+3=(a-1)(a-3).$$
Тогда
$$\frac{2a^{\frac13}}{a^{\frac13}(a-3)} -\frac{a^{\frac23}}{a^{\frac23}(a-1)} -\frac{a+1}{(a-1)(a-3)} = \frac{2}{a-3}-\frac{1}{a-1}-\frac{a+1}{(a-1)(a-3)}.$$
Приведём к общему знаменателю $$\left(a-1\right)\left(a-3\right)$$:
$$\frac{2(a-1)-(a-3)-(a+1)}{(a-1)(a-3)} = \frac{2a-2-a+3-a-1}{(a-1)(a-3)} = \frac{0}{(a-1)(a-3)}=0.$$
Ответ: $$0$$
