Упр.23.17 ГДЗ Мерзляк Поляков 9 класс (Алгебра)
1) (a-5a^(1/2))/(a^(1/2)-5);
2) (a-4b)/(a^0,5+2b^0,5);
3) (a-b)/(ab^(1/2)+a^(1/2) b);
4) (a+2a^(1/2) b^(1/2)+b)/(a^(1/2)+b^(1/2));
5) (4c^(2/3)-12c^(1/3) d^(1/3)+9d^(2/3))/(2c^(1/3)-3d^(1/3));
6) (a+b)/(a^(1/3)+b^(1/3));
7) (m^(1/2)-n^(1/2))/(m^(3/2)-n^(3/2));
8) (a^(3/4)+7a^(1/2))/(a-49a^(1/2));
9) (30^(1/5)-6^(1/5))/(10^(1/5)-2^(1/5)).
$$\frac{a-5a^{1/2}}{a^{1/2}-5}=\frac{a^{1/2}(a^{1/2}-5)}{a^{1/2}-5}=a^{1/2}.$$
$$\frac{a-4b}{a^{0,5}+2b^{0,5}}=\frac{(a^{0,5}-2b^{0,5})(a^{0,5}+2b^{0,5})}{a^{0,5}+2b^{0,5}}=a^{0,5}-2b^{0,5}.$$
$$\frac{a-b}{ab^{1/2}+a^{1/2}b}=\frac{(a^{1/2}-b^{1/2})(a^{1/2}+b^{1/2})}{a^{1/2}b^{1/2}(a^{1/2}+b^{1/2})}=\frac{a^{1/2}-b^{1/2}}{a^{1/2}b^{1/2}}.$$
$$\frac{a^{1/2}-b^{1/2}}{a^{1/2}b^{1/2}}=\frac{1}{b^{1/2}}-\frac{1}{a^{1/2}}.$$
$$\frac{a+2a^{1/2}b^{1/2}+b}{a^{1/2}+b^{1/2}}=\frac{(a^{1/2}+b^{1/2})^2}{a^{1/2}+b^{1/2}}=a^{1/2}+b^{1/2}.$$
$$\frac{4c^{2/3}-12c^{1/3}d^{1/3}+9d^{2/3}}{2c^{1/3}-3d^{1/3}}=\frac{(2c^{1/3}-3d^{1/3})^2}{2c^{1/3}-3d^{1/3}}=2c^{1/3}-3d^{1/3}.$$
$$\frac{a+b}{a^{1/3}+b^{1/3}}=\frac{a^{2/3}+b^{2/3}}{a^{1/3}+b^{1/3}}.$$
$$a+b=(a^{1/3}+b^{1/3})(a^{2/3}-a^{1/3}b^{1/3}+b^{2/3}),$$
значит,
$$\frac{a+b}{a^{1/3}+b^{1/3}}=a^{2/3}-a^{1/3}b^{1/3}+b^{2/3}.$$$$\frac{m^{1/2}-n^{1/2}}{m^{3/2}-n^{3/2}}=\frac{m^{1/2}-n^{1/2}}{(m^{1/2}-n^{1/2})(m+n+m^{1/2}n^{1/2})}.$$
$$\frac{m^{1/2}-n^{1/2}}{m^{3/2}-n^{3/2}}=\frac{1}{m+n+(mn)^{1/2}}.$$
$$\frac{a^{3/4}+7a^{1/2}}{a-49a^{1/2}}=\frac{a^{1/2}(a^{1/4}+7)}{a^{1/2}(a^{1/2}-49)}=\frac{a^{1/4}+7}{a^{1/2}-49}.$$
$$a^{1/2}-49=(a^{1/4}-7)(a^{1/4}+7),$$
поэтому
$$\frac{a^{3/4}+7a^{1/2}}{a-49a^{1/2}}=\frac{1}{a^{1/4}-7}.$$$$\frac{30^{1/5}-6^{1/5}}{10^{1/5}-2^{1/5}}=\frac{6^{1/5}(5^{1/5}-1)}{2^{1/5}(5^{1/5}-1)}=\frac{6^{1/5}}{2^{1/5}}=3^{1/5}.$$
Ответ
1) $$a^{0,5}$$; 2) $$a^{0,5}-2b^{0,5}$$; 3) $$b^{-0,5}-a^{-0,5}$$; 4) $$a^{0,5}+b^{0,5}$$; 5) $$2c^{1/3}-3d^{1/3}$$; 6) $$a^{2/3}-a^{1/3}b^{1/3}+b^{2/3}$$; 7) $$\frac{1}{m+n+(mn)^{0,5}}$$; 8) $$\frac{1}{a^{0,25}-7}$$; 9) $$3^{1/5}$$.
