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

а) (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)·a^(-1/3);
б) (2(x^(1/4)-y^(1/4))/(x^(-1/2)y^(-1/2)-x^(-1/4)y^(-1/2))-x-y):(y-x)/(x^(1/2)-y^(1/2);
в) (c-1)/(c^(3/4)+c^(1/2))·(c^(1/2)+c^(1/4))/(c^(1/2)+1)·c^(1/4)+1;
г) (3(ab)^(1/2)-3b)/(a-b)+((a^(1/2)-b^(1/2))^3+2a^(3/2)+b^(3/2))/(a^(3/2)+b^(3/2)).

1. а)

   $$\frac{a^{7/3}-2a^{5/3}b^{2/3}+ab^{4/3}}{a^{5/3}-a^{4/3}b^{1/3}-ab^{2/3}+a^{2/3}b}\cdot a^{-1/3}$$
   $$=\frac{a\left(a^{4/3}-2a^{2/3}b^{2/3}+b^{4/3}\right)}{a^{2/3}\left(a^{1/3}-b^{1/3}\right)-a^{2/3}b^{1/3}\left(a^{1/3}-b^{1/3}\right)}\cdot a^{-1/3}$$
   $$=\frac{a\left(a^{2/3}-b^{2/3}\right)^2}{a^{2/3}\left(a^{2/3}-b^{2/3}\right)\left(a^{1/3}-b^{1/3}\right)}\cdot a^{-1/3}$$
   $$=\frac{a\left(a^{1/3}-b^{1/3}\right)\left(a^{1/3}+b^{1/3}\right)}{a\left(a^{1/3}-b^{1/3}\right)}=a^{1/3}+b^{1/3}.$$

2. б)

   $$\left(\frac{2\left(x^{1/4}-y^{1/4}\right)}{x^{-1/2}y^{-1/2}-x^{-1/4}y^{-1/2}}-x-y\right):\frac{y-x}{x^{1/2}-y^{1/2}}$$
   $$=\left(\frac{2\left(x^{1/4}-y^{1/4}\right)}{x^{-1/2}y^{-1/2}\left(1-x^{1/4}y^{1/4}\right)}-x-y\right)\cdot\frac{x^{1/2}-y^{1/2}}{y-x}$$
   $$=\left(-2x^{1/2}y^{1/2}-x-y\right)\cdot\frac{x^{1/2}-y^{1/2}}{y-x}$$
   $$=-\left(x^{1/2}+y^{1/2}\right)^2\cdot\frac{x^{1/2}-y^{1/2}}{y-x}.$$
   Так как $$y-x=-(x^{1/2}-y^{1/2})(x^{1/2}+y^{1/2}),$$ то
   $$-\left(x^{1/2}+y^{1/2}\right)^2\cdot\frac{x^{1/2}-y^{1/2}}{y-x}=x^{1/2}+y^{1/2}.$$

3. в)

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

4. г)

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

Ответ

а) $$a^{1/3}+b^{1/3}$$; б) $$x^{1/2}+y^{1/2}$$; в) $$c^{1/2}$$; г) $$3$$.