Упр.743 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) (2+v2+v3)/(v2+v3+v6+v8+4);
2) (2-v3)/(v2-v(2-v3))+(2+v3)/(v2+v(2+v3));
3) (v(x-2v(x-1))+v(x+2v(x-1)))/(v(x^2-4(x-1))), если 1 < x < 2; 4) ((a+b)/((a^2)^(1/3)-(b^2)^(1/3))+((ab^2)^(1/3)-(a^2 b)^(1/3))/((a^2)^(1/3)-2(ab)^(1/3)+(b^2)^(1/3))):(a^(1/6)-b^(1/6))-b^(1/6).

1. $$\frac{2+\sqrt2+\sqrt3}{\sqrt2+\sqrt3+\sqrt6+\sqrt8+4}$$
   $$=\frac{2+\sqrt2+\sqrt3}{\sqrt2+\sqrt3+\sqrt{2\cdot 3}+2\sqrt2+2+\sqrt{2\cdot 2}}$$
   $$=\frac{2+\sqrt2+\sqrt3}{(\sqrt2+\sqrt3+2)+\sqrt2(\sqrt2+\sqrt3+2)}$$
   $$=\frac{2+\sqrt2+\sqrt3}{(1+\sqrt2)(\sqrt2+\sqrt3+2)}=\frac{1}{1+\sqrt2}$$
   $$=\frac{1-\sqrt2}{(1+\sqrt2)(1-\sqrt2)}=\frac{1-\sqrt2}{1-2}=\sqrt2-1.$$

2. $$\frac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}+\frac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}$$
   $$=\frac{\sqrt2(2-\sqrt3)}{\sqrt2\bigl(\sqrt2-\sqrt{2-\sqrt3}\bigr)}+\frac{\sqrt2(2+\sqrt3)}{\sqrt2\bigl(\sqrt2+\sqrt{2+\sqrt3}\bigr)}$$
   $$=\frac{\sqrt2(2-\sqrt3)}{2-\sqrt{4-2\sqrt3}}+\frac{\sqrt2(2+\sqrt3)}{2+\sqrt{4+2\sqrt3}}$$
   $$=\frac{\sqrt2(2-\sqrt3)}{2-\sqrt{(\sqrt3-1)^2}}+\frac{\sqrt2(2+\sqrt3)}{2+\sqrt{(\sqrt3+1)^2}}$$
   $$=\frac{\sqrt2(2-\sqrt3)}{2-(\sqrt3-1)}+\frac{\sqrt2(2+\sqrt3)}{2+(\sqrt3+1)}$$
   $$=\frac{\sqrt2(2-\sqrt3)}{3-\sqrt3}+\frac{\sqrt2(2+\sqrt3)}{3+\sqrt3}$$
   $$=\frac{(2\sqrt2-\sqrt6)(3+\sqrt3)+(2\sqrt2+\sqrt6)(3-\sqrt3)}{(3-\sqrt3)(3+\sqrt3)}$$
   $$=\frac{6\sqrt2+2\sqrt6-3\sqrt6-\sqrt{18}+6\sqrt2-2\sqrt6+3\sqrt6-\sqrt{18}}{9-3}$$
   $$=\frac{12\sqrt2-2\sqrt{18}}{6}=\frac{12\sqrt2-6\sqrt2}{6}=\sqrt2.$$

3. $$\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4(x-1)}}$$
   $$=\frac{\sqrt{(\sqrt{x-1})^2-2\sqrt{x-1}+1}+\sqrt{(\sqrt{x-1})^2+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}$$
   $$=\frac{(1-\sqrt{x-1})+(1+\sqrt{x-1})}{2-x}=\frac{2}{2-x}, \quad 1

4. $$\left(\frac{a+b}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}+\frac{\sqrt[3]{ab^2}-\sqrt[3]{a^2b}}{\sqrt[3]{a^2}-2\sqrt[3]{ab}+\sqrt[3]{b^2}}\right):(\sqrt[6]{a}-\sqrt[6]{b})-\sqrt[6]{b}$$
   $$=\left(\frac{a+b}{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a}+\sqrt[3]{b})}+\frac{\sqrt[3]{ab}(\sqrt[3]{b}-\sqrt[3]{a})}{(\sqrt[3]{a}-\sqrt[3]{b})^2}\right):(\sqrt[6]{a}-\sqrt[6]{b})-\sqrt[6]{b}$$
   $$=\frac{(a+b)-\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})}{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a}+\sqrt[3]{b})}:(\sqrt[6]{a}-\sqrt[6]{b})-\sqrt[6]{b}$$
   $$=\frac{(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[3]{a}-\sqrt[3]{ab}+\sqrt[3]{b})-\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})}{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[6]{a}-\sqrt[6]{b})}-\sqrt[6]{b}$$
   $$=\frac{\sqrt[3]{a}-2\sqrt[3]{ab}+\sqrt[3]{b}}{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[6]{a}-\sqrt[6]{b})}-\sqrt[6]{b}$$
   $$=\frac{(\sqrt[6]{a}-\sqrt[6]{b})^2}{(\sqrt[3]{a}-\sqrt[3]{b})(\sqrt[6]{a}-\sqrt[6]{b})}-\sqrt[6]{b}$$
   $$=\frac{\sqrt[6]{a}-\sqrt[6]{b}}{\sqrt[3]{a}-\sqrt[3]{b}}-\sqrt[6]{b}=\sqrt[6]{a}+\sqrt[6]{b}-\sqrt[6]{b}=\sqrt[6]{a}.$$

Ответ

1) $$\sqrt2-1$$;
2) $$\sqrt2$$;
3) $$\frac{2}{2-x}$$;
4) $$\sqrt[6]{a}$$.