Упр.10.25 ГДЗ Мерзляк 10 класс Базовый уровень (Алгебра)

1) (a-b)/(a^0,5-b^0,5)-(a^1,5-b^1,5)/(a-b);
2) (a^0,5-b^0,5)/(a^0,5+b^0,5)+(a^0,5+b^0,5)/(a^0,5-b^0,5);
3) (a^(1/2)+2a^(1/4) b^(1/4)+b^(1/2))/(a^(7/6) b^(5/6)-a^(5/6) b^(7/6))·(a-a^(1/3) b^(2/3))/(a^(1/4) b^(1/4)+b^(1/2));
4) (a^(3/2)+b^(3/2))/(a^2-ab)^(2/3):(a^(-2/3)(a-b)^(1/3))/(a^(3/2)-b^(3/2).

1. $$\frac{a-b}{a^{0,5}-b^{0,5}}-\frac{a^{1,5}-b^{1,5}}{a-b}$$
   Приведём к общему знаменателю:
   $$\frac{(a-b)(a^{0,5}+b^{0,5})-(a^{1,5}-b^{1,5})}{(a^{0,5}-b^{0,5})(a^{0,5}+b^{0,5})}$$
   Так как
   $$a-b=(a^{0,5}-b^{0,5})(a^{0,5}+b^{0,5}),$$
   то числитель равен
   $$a^{1,5}+ab^{0,5}-ba^{0,5}-b^{1,5}-a^{1,5}+b^{1,5}=ab^{0,5}-ba^{0,5}.$$
   Тогда
   $$\frac{ab^{0,5}-ba^{0,5}}{(a^{0,5}-b^{0,5})(a^{0,5}+b^{0,5})} =\frac{a^{0,5}b^{0,5}(a^{0,5}-b^{0,5})}{(a^{0,5}-b^{0,5})(a^{0,5}+b^{0,5})} =\frac{a^{0,5}b^{0,5}}{a^{0,5}+b^{0,5}}.$$

2. $$\frac{a^{0,5}-b^{0,5}}{a^{0,5}+b^{0,5}}+\frac{a^{0,5}+b^{0,5}}{a^{0,5}-b^{0,5}}$$
   Приведём к общему знаменателю:
   $$\frac{(a^{0,5}-b^{0,5})^2+(a^{0,5}+b^{0,5})^2}{(a^{0,5}+b^{0,5})(a^{0,5}-b^{0,5})}$$
   $$=\frac{a-2a^{0,5}b^{0,5}+b+a+2a^{0,5}b^{0,5}+b}{a-b} =\frac{2a+2b}{a-b} =\frac{2(a+b)}{a-b}.$$

3. $$\frac{a^{1/2}+2a^{1/4}b^{1/4}+b^{1/2}}{a^{7/6}b^{5/6}-a^{5/6}b^{7/6}}\cdot \frac{a-a^{1/3}b^{2/3}}{a^{1/4}b^{1/4}+b^{1/2}}$$
   Разложим на множители:
   $$a^{1/2}+2a^{1/4}b^{1/4}+b^{1/2}=(a^{1/4}+b^{1/4})^2,$$
   $$a^{7/6}b^{5/6}-a^{5/6}b^{7/6}=a^{5/6}b^{5/6}(a^{1/3}-b^{1/3}),$$
   $$a-a^{1/3}b^{2/3}=a^{1/3}(a^{2/3}-b^{2/3})=a^{1/3}(a^{1/3}-b^{1/3})(a^{1/3}+b^{1/3}),$$
   $$a^{1/4}b^{1/4}+b^{1/2}=b^{1/4}(a^{1/4}+b^{1/4}).$$
   Тогда
   $$\frac{(a^{1/4}+b^{1/4})^2}{a^{5/6}b^{5/6}(a^{1/3}-b^{1/3})}\cdot \frac{a^{1/3}(a^{1/3}-b^{1/3})(a^{1/3}+b^{1/3})}{b^{1/4}(a^{1/4}+b^{1/4})}$$
   $$=\frac{(a^{1/4}+b^{1/4})(a^{1/3}+b^{1/3})}{a^{1/2}b^{13/12}}.$$

4. $$\frac{a^{3/2}+b^{3/2}}{(a^2-ab)^{2/3}}:\frac{a^{-2/3}(a-b)^{1/3}}{a^{3/2}-b^{3/2}}$$
   Заменим деление умножением на обратную дробь:
   $$\frac{a^{3/2}+b^{3/2}}{(a^2-ab)^{2/3}}\cdot\frac{a^{3/2}-b^{3/2}}{a^{-2/3}(a-b)^{1/3}}.$$
   Так как
   $$a^2-ab=a(a-b),$$
   то
   $$(a^2-ab)^{2/3}=a^{2/3}(a-b)^{2/3}.$$
   Кроме того,
   $$a^{3/2}+b^{3/2}=(a^{1/2}+b^{1/2})(a+b-\sqrt{ab}),$$
   но удобнее использовать формулу
   $$a^{3/2}-b^{3/2}=(a^{1/2}-b^{1/2})(a+b+\sqrt{ab}).$$
   После сокращения степеней получаем:
   $$\frac{a^{3/2}+b^{3/2}}{a^{2/3}(a-b)^{2/3}}\cdot\frac{a^{2/3}(a^{3/2}-b^{3/2})}{(a-b)^{1/3}} =\frac{a^{3/2}+b^{3/2}}{(a-b)}\cdot(a^{3/2}-b^{3/2}).$$
   Тогда
   $$\frac{(a^{3/2}+b^{3/2})(a^{3/2}-b^{3/2})}{a-b} =\frac{a^3-b^3}{a-b} =a^2+ab+b^2.$$

Ответ

1) $$\frac{a^{0,5}b^{0,5}}{a^{0,5}+b^{0,5}}$$

2) $$\frac{2(a+b)}{a-b}$$

3) $$\frac{(a^{1/4}+b^{1/4})(a^{1/3}+b^{1/3})}{a^{1/2}b^{13/12}}$$

4) $$a^2+ab+b^2$$