Упр.569 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

Задача

1) (m^(1/2) n^(1/5)+n^(1/3))/(m+m^(1/2) n^(2/15)) при m=0,04, n=243;
2) (m^(1/2) n^(-2)-n^(-3/2))/(m^(3/4)-m^(1/4) n^(1/2)) при m=81, n=0,1;
3) (1+v((a-x)/(a+x)))·(1-v((a-x)/(a+x))) при a=5, x=4;
4) (a+v(a^2-x^2))/(a-v(a^2-x^2))-(a-v(a^2-x^2))/(a+v(a^2-x^2)) при a=3, x=v5.

Подробный ответ

$$\frac{m^{1/2}n^{1/5}+n^{1/3}}{m+m^{1/2}n^{2/15}}= \frac{n^{1/5}\left(m^{1/2}+n^{2/15}\right)}{m^{1/2}\left(m^{1/2}+n^{2/15}\right)}= \frac{n^{1/5}}{m^{1/2}}= \frac{\sqrt[5]{n}}{\sqrt{m}}.$$
При $$m=0{,}04,\ n=243$$ получаем:
$$\frac{\sqrt[5]{243}}{\sqrt{0{,}04}}=\frac{3}{0{,}2}=15.$$
$$\frac{m^{1/2}n^{-2}-n^{-3/2}}{m^{3/4}-m^{1/4}n^{1/2}}= \frac{n^{-2}\left(m^{1/2}-n^{1/2}\right)}{m^{1/4}\left(m^{1/2}-n^{1/2}\right)}= \frac{n^{-2}}{m^{1/4}}= \frac{1}{n^2\sqrt[4]{m}}.$$
При $$m=81,\ n=0{,}1$$:
$$\frac{1}{n^2\sqrt[4]{m}}=\frac{1}{0{,}1^2\cdot \sqrt[4]{81}}=\frac{1}{0{,}01\cdot 3}=\frac{100}{3}=33\frac{1}{3}.$$
$$\left(1+\sqrt{\frac{a-x}{a+x}}\right)\left(1-\sqrt{\frac{a-x}{a+x}}\right) =1-\frac{a-x}{a+x} =\frac{(a+x)-(a-x)}{a+x} =\frac{2x}{a+x}.$$
При $$a=5,\ x=4$$:
$$\frac{2x}{a+x}=\frac{2\cdot 4}{5+4}=\frac{8}{9}.$$
$$\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}= \frac{\left(a+\sqrt{a^2-x^2}\right)^2-\left(a-\sqrt{a^2-x^2}\right)^2}{a^2-\left(a^2-x^2\right)}.$$
Числитель:
$$\left(a+\sqrt{a^2-x^2}\right)^2-\left(a-\sqrt{a^2-x^2}\right)^2=4a\sqrt{a^2-x^2}.$$
Знаменатель:
$$a^2-\left(a^2-x^2\right)=x^2.$$
Значит,
$$\frac{a+\sqrt{a^2-x^2}}{a-\sqrt{a^2-x^2}}-\frac{a-\sqrt{a^2-x^2}}{a+\sqrt{a^2-x^2}}= \frac{4a\sqrt{a^2-x^2}}{x^2}.$$
При $$a=3,\ x=\sqrt{5}$$:
$$\frac{4\cdot 3\cdot \sqrt{3^2-(\sqrt{5})^2}}{(\sqrt{5})^2} =\frac{12\sqrt{9-5}}{5} =\frac{12\cdot 2}{5} =\frac{24}{5}=4{,}8.$$