Упр.743 ГДЗ Колягин Ткачёва 9 класс (Алгебра)
- Упростить выражение:
а) $$\frac{2+\sqrt{2}+\sqrt{3}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4};$$
б) $$\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}+\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}};$$
в) $$\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4(x-1)}}$$, если $$1<x<2$$;
г) $$\left(\frac{a+b}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}+\frac{\sqrt[3]{ab^2}-\sqrt[3]{a^2b}}{\sqrt[3]{a^2}-2\sqrt[3]{ab}+\sqrt[3]{b^2}}\right):\left(\sqrt[6]{a}-\sqrt[6]{b}\right)-\sqrt[6]{b}.$$
$$\frac{2+\sqrt2+\sqrt3}{\sqrt2+\sqrt3+\sqrt6+\sqrt8+4}$$
$$=\frac{2+\sqrt2+\sqrt3}{\sqrt2+\sqrt3+\sqrt{2\cdot 3}+2\sqrt2+2+\sqrt{2\cdot 2}}$$
$$=\frac{2+\sqrt2+\sqrt3}{(\sqrt2+\sqrt3+2)+\sqrt2(\sqrt2+\sqrt3+2)}$$
$$=\frac{2+\sqrt2+\sqrt3}{(1+\sqrt2)(\sqrt2+\sqrt3+2)}=\frac{1}{1+\sqrt2}$$
$$=\frac{1-\sqrt2}{(1+\sqrt2)(1-\sqrt2)}=\frac{1-\sqrt2}{1-2}=\sqrt2-1.$$$$\frac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}+\frac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}$$
$$=\frac{\sqrt2(2-\sqrt3)}{\sqrt2\bigl(\sqrt2-\sqrt{2-\sqrt3}\bigr)}+\frac{\sqrt2(2+\sqrt3)}{\sqrt2\bigl(\sqrt2+\sqrt{2+\sqrt3}\bigr)}$$
$$=\frac{\sqrt2(2-\sqrt3)}{2-\sqrt{4-2\sqrt3}}+\frac{\sqrt2(2+\sqrt3)}{2+\sqrt{4+2\sqrt3}}$$
$$=\frac{\sqrt2(2-\sqrt3)}{2-\sqrt{(\sqrt3-1)^2}}+\frac{\sqrt2(2+\sqrt3)}{2+\sqrt{(\sqrt3+1)^2}}$$
$$=\frac{\sqrt2(2-\sqrt3)}{2-(\sqrt3-1)}+\frac{\sqrt2(2+\sqrt3)}{2+(\sqrt3+1)}$$
$$=\frac{\sqrt2(2-\sqrt3)}{3-\sqrt3}+\frac{\sqrt2(2+\sqrt3)}{3+\sqrt3}$$
$$=\frac{(2\sqrt2-\sqrt6)(3+\sqrt3)+(2\sqrt2+\sqrt6)(3-\sqrt3)}{(3-\sqrt3)(3+\sqrt3)}$$
$$=\frac{6\sqrt2+2\sqrt6-3\sqrt6-\sqrt{18}+6\sqrt2-2\sqrt6+3\sqrt6-\sqrt{18}}{9-3}$$
$$=\frac{12\sqrt2-2\sqrt{18}}{6}=\frac{12\sqrt2-6\sqrt2}{6}=\sqrt2.$$$$\frac{\sqrt{x-2\sqrt{x-1}}+\sqrt{x+2\sqrt{x-1}}}{\sqrt{x^2-4(x-1)}}$$
$$=\frac{\sqrt{(\sqrt{x-1})^2-2\sqrt{x-1}+1}+\sqrt{(\sqrt{x-1})^2+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}$$
$$=\frac{(1-\sqrt{x-1})+(1+\sqrt{x-1})}{2-x}=\frac{2}{2-x}, \quad 1$$\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}$$.
