Упр.570 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) 6n·v(m/2n)·v(18mn);
2) (m+n)/(m+2vmn+n):((vm+vn)/(vm-vn)-2vmn/(m-n));
3) (a-1)/(a^(3/4)+a^(1/2))·(a^(1/2)+a^(1/4))/(a^(1/2)+1)·a^(1/4);
4) (4/(4-a)+(2-a^(1/2))/(2a^(1/2)+a)):(16+8a+a^2)/a^(3/2).

1. $$6n\cdot \sqrt{\frac{m}{2n}}\cdot \sqrt{18mn}=6n\cdot \sqrt{9m^2}=6n\cdot 3|m|=18n|m|.$$
2. $$\frac{m+n}{m+2\sqrt{mn}+n}:\left(\frac{\sqrt m+\sqrt n}{\sqrt m-\sqrt n}-\frac{2\sqrt{mn}}{m-n}\right)$$
   $$=\frac{m+n}{(\sqrt m+\sqrt n)^2}:\frac{(\sqrt m+\sqrt n)^2-2\sqrt{mn}}{(\sqrt m-\sqrt n)(\sqrt m+\sqrt n)}$$
   $$=\frac{m+n}{(\sqrt m+\sqrt n)^2}:\frac{m+n+2\sqrt{mn}-2\sqrt{mn}}{(\sqrt m-\sqrt n)(\sqrt m+\sqrt n)}$$
   $$=\frac{m+n}{(\sqrt m+\sqrt n)^2}:\frac{m+n}{m-n}$$
   $$=\frac{m+n}{(\sqrt m+\sqrt n)^2}\cdot \frac{m-n}{m+n}=\frac{m-n}{(\sqrt m+\sqrt n)^2}=\frac{\sqrt m-\sqrt n}{\sqrt m+\sqrt n}.$$
3. $$\frac{a-1}{a^{3/4}+a^{1/2}}\cdot \frac{a^{1/2}+a^{1/4}}{a^{1/2}+1}\cdot a^{1/4}$$
   $$=\frac{(a^{1/2}-1)(a^{1/2}+1)}{a^{1/2}(a^{1/4}+1)}\cdot \frac{a^{1/2}+a^{1/4}}{a^{1/2}+1}\cdot a^{1/4}$$
   $$=\frac{a^{1/2}-1}{a^{1/2}}\cdot \frac{a^{1/2}+a^{1/4}}{a^{1/4}+1}\cdot a^{1/4}$$
   $$=\frac{a^{1/2}-1}{a^{1/2}}\cdot \frac{a^{1/4}(a^{1/4}+1)}{a^{1/4}+1}\cdot a^{1/4}$$
   $$=a^{1/2}-1=\sqrt a-1.$$
4. $$\left(\frac{4}{4-a}+\frac{2-\sqrt a}{2\sqrt a+a}\right):\frac{16+8a+a^2}{a^{3/2}}$$
   $$=\left(\frac{4}{(2-\sqrt a)(2+\sqrt a)}+\frac{2-\sqrt a}{\sqrt a(2+\sqrt a)}\right):\frac{(4+a)^2}{a^{3/2}}$$
   $$=\left(\frac{4\sqrt a+(2-\sqrt a)^2}{\sqrt a(2-\sqrt a)(2+\sqrt a)}\right)\cdot \frac{a^{3/2}}{(4+a)^2}$$
   $$=\frac{4\sqrt a+4-4\sqrt a+a}{\sqrt a(2-\sqrt a)(2+\sqrt a)}\cdot \frac{a^{3/2}}{(4+a)^2}$$
   $$=\frac{a+4}{\sqrt a(4-a)}\cdot \frac{a^{3/2}}{(4+a)^2}=\frac{a}{(4-a)(4+a)}=\frac{a}{16-a^2}.$$

Ответ

1) $$18n|m|$$; 2) $$\frac{\sqrt m-\sqrt n}{\sqrt m+\sqrt n}$$; 3) $$\sqrt a-1$$; 4) $$\frac{a}{16-a^2}$$.