Сандық өрнектерді тепе-тең түрлендіру (Рустюмова 1.1.5)

Шешуі: $$(3-\sqrt{2})\sqrt{11+2\cdot 3\cdot \sqrt{2}}=(3+\sqrt{2})\sqrt{3^2+2\cdot 3\sqrt{2}+{(\sqrt{2})}^2}=$$ $$=(3-\sqrt{2})\sqrt{{(3+\sqrt{2})}^2}=(3-\sqrt{2})(3+\sqrt{2})=9-2=7$$

Шешуі: $$(3-\sqrt{5})\sqrt{14+2\cdot 3\cdot \sqrt{5}}=(3-\sqrt{5})\sqrt{3^2+2\cdot 3\cdot \sqrt{5}+{(\sqrt{5})}^2}=$$ $$=(3-\sqrt{5})\sqrt{{(3+\sqrt{5})}^2}=(3-\sqrt{5})(3+\sqrt{5})=9-5=4$$

Шешуі: $$\left(7+\sqrt{33}\right)+2\sqrt{\left(7+\sqrt{33}\right)\left(7-\sqrt{33}\right)}+\left(7-\sqrt{33}\right)=$$ $$=7+\sqrt{33}+8+7-\sqrt{33}=22$$

Шешуі: $$\sqrt{3-2\sqrt{2}}\cdot \sqrt[4]{17+2\cdot 6\sqrt{2}}=$$ $$=\sqrt{3-2\sqrt{2}}\cdot \sqrt[4]{3^2+2\cdot 3\cdot 2\sqrt{2}+{(2\sqrt{2})}^2}=$$ $$=\sqrt{3-2\sqrt{2}}\cdot \sqrt{3+2\sqrt{2}}=\sqrt{9-8}=1$$

Шешуі: $$=\left|\sqrt{3}-1\right|+\left|\sqrt{3}-2\right|=\sqrt{3}-1+2-\sqrt{3}=1$$

Шешуі: $$=\left|2-\sqrt{5}\right|+\left|3-\sqrt{5}\right|=\sqrt{5}-2+3-\sqrt{5}=1$$

Шешуі: $$=2-\sqrt{3}-\left(\sqrt{3}-1\right)+\sqrt{3}=$$ $$=2-\sqrt{3}-\sqrt{3}+1+\sqrt{3}=3-\sqrt{3}$$

Шешуі: $$\sqrt[3]{\sqrt{4-2\sqrt{3}}}\cdot \sqrt[3]{1+\sqrt{3}}\cdot \sqrt[3]{4}=$$ $$=\sqrt[3]{\sqrt{{(\sqrt{3})}^2-2\cdot \sqrt{3}\cdot 1+1^2}}\cdot \sqrt[3]{1+\sqrt{3}}\cdot \sqrt[3]{4}=$$ $$=\sqrt[3]{(\sqrt{3}-1)(\sqrt{3}+1)}\cdot \sqrt[3]{4}=\sqrt[3]{2}\cdot \sqrt[3]{4}=\sqrt[3]{2\cdot 4}=2$$

Шешуі: $$=|-\sqrt{14}+2|+2\sqrt{14}-8=\sqrt{14}-2+2\sqrt{14}-8=3\sqrt{14}-10$$

Шешуі: $$={\displaystyle \frac{\sqrt{5}(5\sqrt{5}+8)}{8+5\sqrt{5}}}\cdot \sqrt{9-2\cdot 2\cdot \sqrt{5}}=\sqrt{5}\cdot \sqrt{2^2-2\cdot 2\cdot \sqrt{5}+{\left(\sqrt{5}\right)}^2}=$$ $$=\sqrt{5}\cdot \sqrt{{(2-\sqrt{5})}^2}=\sqrt{5}\cdot \left(\sqrt{5}-2\right)=5-2\sqrt{5}$$

Шешуі: $$\sqrt[6]{7+4\sqrt{3}}=\sqrt[6]{2^2+2\cdot 2\cdot \sqrt{3}+{\left(\sqrt{3}\right)}^2}=\sqrt[3]{2+\sqrt{3}}$$ $$\sqrt[3]{2-\sqrt{3}}\cdot \sqrt[3]{2+\sqrt{3}}=\sqrt[3]{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=1$$

Шешуі: $$=3+2\sqrt{3}+1+3-4\sqrt{3}+4+3\sqrt{3}+\sqrt{{(\sqrt{3})}^2+2\sqrt{3}+1^2}+\sqrt{2^2-2\cdot 2\cdot \sqrt{3}+{(\sqrt{3})}^2}=$$ $$=11+\sqrt{3}+\sqrt{3}+1+2-\sqrt{3}=14+\sqrt{3}$$

Шешуі: $$2+6\sqrt{2}+9+2-2\sqrt{2}+1+2\sqrt{2}+\sqrt{{(\sqrt{2})}^2+2\cdot \sqrt{2}\cdot 1+1^2}+\sqrt{3^2-2\cdot 3\cdot \sqrt{2}+{(\sqrt{2})}^2}=$$ $$=14+6\sqrt{2}+\sqrt{2}+1+3-\sqrt{2}=18+6\sqrt{2}$$

Шешуі: $$\sqrt{17-4\sqrt{2^2+2\cdot 2\cdot \sqrt{5}+{(\sqrt{5})}^2}}=\sqrt{17-4(2+\sqrt{5})}=\sqrt{17-8-4\sqrt{5}}=$$ $$=\sqrt{9-4\sqrt{5}}=\sqrt{2^2-2\cdot 2\cdot \sqrt{5}+{(\sqrt{5})}^2}=\sqrt{{(2-\sqrt{5})}^2}=\sqrt{5}-2$$

Шешуі: $$\sqrt{15-4\sqrt{2^2+2\cdot 2\cdot \sqrt{3}+{(\sqrt{3})}^2}}=\sqrt{15-4(2+\sqrt{3})}=$$ $$=\sqrt{15-8-4\sqrt{3}}=\sqrt{7-4\sqrt{3}}=\sqrt{2^2-2\cdot 2\cdot \sqrt{3}+{(\sqrt{3})}^2}=$$ $$=\sqrt{{(2-\sqrt{3})}^2}=2-\sqrt{3}$$