Упр.51 Повторение ГДЗ Колмогоров 10-11 класс (Алгебра)

Задача

Упростите выражения:
а) $$\frac{a^{7/3}-2a^{5/3}b^{2/3}+ab^{4/3}}{a^{5/3}-a^{4/3}b^{1/3}-ab^{2/3}+a^{2/3}b}\cdot a^{-1/3}$$;
б) $$\left(\frac{2\left(x^{1/4}-y^{1/4}\right)}{x^{-1/2}y^{-1/2}-x^{-1/4}y^{-1/2}}-x-y\right):\frac{y-x}{x^{1/2}-y^{1/2}}$$;
в) $$\frac{c-1}{c^{3/4}+c^{1/2}}\cdot\frac{c^{1/2}+c^{1/4}}{c^{1/2}+1}\cdot c^{1/4}+1$$;
г) $$\frac{3(ab)^{1/2}-3b}{a-b}+\frac{\left(a^{1/2}-b^{1/2}\right)^3+2a^{3/2}+b^{3/2}}{a^{3/2}+b^{3/2}}$$.

Подробный ответ

а) $$\frac{a^{7/3}-2a^{5/3}b^{2/3}+ab^{4/3}}{a^{5/3}-a^{4/3}b^{1/3}-ab^{2/3}+a^{2/3}b}\cdot a^{-1/3}$$

Вынесем общий множитель и разложим на множители:

$$a^{7/3}-2a^{5/3}b^{2/3}+ab^{4/3}=a\left(a^{2/3}-b^{2/3}\right)^2,$$

$$a^{5/3}-a^{4/3}b^{1/3}-ab^{2/3}+a^{2/3}b=a^{2/3}\left(a^{1/3}-b^{1/3}\right)\left(a^{1/3}-b^{1/3}\right)=a^{2/3}\left(a^{1/3}-b^{1/3}\right)^2.$$

Тогда

$$\frac{a\left(a^{2/3}-b^{2/3}\right)^2}{a^{2/3}\left(a^{1/3}-b^{1/3}\right)^2}\cdot a^{-1/3} =\frac{a\left(a^{1/3}-b^{1/3}\right)^2\left(a^{1/3}+b^{1/3}\right)^2}{a^{2/3}\left(a^{1/3}-b^{1/3}\right)^2}\cdot a^{-1/3} =a^{1/3}+b^{1/3}.$$

б) $$\left(\frac{2\left(x^{1/4}-y^{1/4}\right)}{x^{-1/2}y^{-1/4}-x^{-1/4}y^{-1/2}}-x-y\right):\frac{y-x}{x^{1/2}-y^{1/2}}$$

Преобразуем дробь в скобках:

$$x^{-1/2}y^{-1/4}-x^{-1/4}y^{-1/2}=x^{-1/2}y^{-1/2}\left(y^{1/4}-x^{1/4}\right),$$

поэтому

$$\frac{2\left(x^{1/4}-y^{1/4}\right)}{x^{-1/2}y^{-1/4}-x^{-1/4}y^{-1/2}} =-2x^{1/2}y^{1/2}.$$

Тогда

$$-2x^{1/2}y^{1/2}-x-y=-(x^{1/2}+y^{1/2})^2.$$

Кроме того,

$$\frac{y-x}{x^{1/2}-y^{1/2}}=\frac{(y^{1/2}-x^{1/2})(y^{1/2}+x^{1/2})}{x^{1/2}-y^{1/2}}=-(x^{1/2}+y^{1/2}).$$

Следовательно,

$$\frac{-(x^{1/2}+y^{1/2})^2}{-(x^{1/2}+y^{1/2})}=x^{1/2}+y^{1/2}.$$

в) $$\frac{c-1}{c^{3/4}+c^{1/2}}\cdot\frac{c^{1/2}+c^{1/4}}{c^{1/2}+1}\cdot c^{1/4}+1$$

Вынесем степени:

$$c^{3/4}+c^{1/2}=c^{1/2}\left(c^{1/4}+1\right), \qquad c^{1/2}+c^{1/4}=c^{1/4}\left(c^{1/4}+1\right).$$

Тогда

$$\frac{c-1}{c^{1/2}\left(c^{1/4}+1\right)}\cdot\frac{c^{1/4}\left(c^{1/4}+1\right)}{c^{1/2}+1}\cdot c^{1/4}+1 =\frac{c^{1/2}(c-1)}{c^{1/2}(c^{1/2}+1)}+1.$$

Так как $$c-1=\left(c^{1/2}-1\right)\left(c^{1/2}+1\right),$$ получаем

$$\frac{c^{1/2}-1}{1}+1=c^{1/2}.$$

г) $$\frac{3(ab)^{1/2}-3b}{a-b}+\frac{\left(a^{1/2}-b^{1/2}\right)^3+2a^{3/2}+b^{3/2}}{a^{3/2}+b^{3/2}}$$

Первую дробь преобразуем:

$$\frac{3b^{1/2}\left(a^{1/2}-b^{1/2}\right)}{\left(a^{1/2}-b^{1/2}\right)\left(a^{1/2}+b^{1/2}\right)}=\frac{3b^{1/2}}{a^{1/2}+b^{1/2}}.$$

Во второй дроби раскроем куб:

$$\left(a^{1/2}-b^{1/2}\right)^3=a^{3/2}-3ab^{1/2}+3a^{1/2}b-b^{3/2}.$$

Тогда числитель второй дроби равен

$$a^{3/2}-3ab^{1/2}+3a^{1/2}b-b^{3/2}+2a^{3/2}+b^{3/2}=3a^{3/2}-3ab^{1/2}+3a^{1/2}b.$$

Следовательно,

$$\frac{3a^{3/2}-3ab^{1/2}+3a^{1/2}b}{a^{3/2}+b^{3/2}} =\frac{3a^{1/2}\left(a-b^{1/2}a^{1/2}+b\right)}{a^{3/2}+b^{3/2}} =\frac{3a^{1/2}}{a^{1/2}+b^{1/2}}.$$

Тогда

$$\frac{3b^{1/2}}{a^{1/2}+b^{1/2}}+\frac{3a^{1/2}}{a^{1/2}+b^{1/2}}=3.$$