Құрамында радикалы бар өрнектерді түрлендіру

$$\sqrt[3]{{2\sqrt {2 \cdot \sqrt[3]{2}} }} = \sqrt[3]{{2\sqrt {\sqrt[3]{{{2^3}}} \cdot \sqrt[3]{2}} }} = \sqrt[3]{{2\sqrt {\sqrt[3]{{{2^3} \cdot 2}}} }} = \sqrt[3]{{2\sqrt {\sqrt[3]{{{2^4}}}} }} = $$ $$ = \sqrt[3]{{2 \cdot \sqrt[6]{{{2^4}}}}} = = \sqrt[3]{{2 \cdot \sqrt[3]{{{2^2}}}}} = \sqrt[3]{{\sqrt[3]{{{2^3} \cdot {2^2}}}}} = \sqrt[3]{{\sqrt[3]{{{2^5}}}}} = \sqrt[9]{{{2^5}}} = \sqrt[9]{{32}}$$

Радикалдарды бірдей дәрежелі түбірге келтіреміз:

$$\frac{{\sqrt[3]{3} \cdot \sqrt[4]{2}}}{{\sqrt[6]{6}}} = \frac{{\sqrt[{12}]{{{3^4}}} \cdot \sqrt[{12}]{{{2^3}}}}}{{\sqrt[{12}]{{{6^2}}}}} = \sqrt[{12}]{{\frac{{{3^4} \cdot {2^3}}}{{{3^2} \cdot {2^2}}}}} = \sqrt[{12}]{{{3^2} \cdot 2}} = \sqrt[{12}]{{18}}$$

$$\sqrt[3]{3} \cdot \sqrt[3]{{27}} \cdot \sqrt[3]{9} – \sqrt[5]{2}:\sqrt[5]{{ – 64}} = 3 \cdot \sqrt[3]{{27}} – \sqrt[5]{{ – \frac{2}{{64}}}} = 9 – \sqrt[5]{{ – \frac{1}{{32}}}} = 9 – \left( { – \frac{1}{2}} \right) = 9,5$$

Түбір астындағы өрнектерден санның квадраттарын көбейткіш ретінде бөліп алып, түбірден шығарамыз, ал бөлшектердің бөліміндегі иррационалдықтан құтыламыз.

$$\sqrt {150} – \sqrt {96} – \sqrt {\frac{2}{3}} – \frac{1}{{\sqrt 6 }} = \sqrt {25 \cdot 6} – \sqrt {16 \cdot 6} – $$ $$ – \frac{{\sqrt 6 }}{3} – \frac{{\sqrt 6 }}{6} = 5\sqrt 6 – 4\sqrt 6 – \frac{1}{2}\sqrt 6 = \frac{{\sqrt 6 }}{2}$$

$$2\sqrt {32} – \frac{1}{3}\sqrt {18} – \frac{1}{2}\sqrt {50} – \frac{1}{2}\sqrt 2 + 3\sqrt 8 = $$ $$ = 8\sqrt 2 – \sqrt 2 – \frac{5}{2}\sqrt 2 – \frac{1}{2}\sqrt 2 + 6\sqrt 2 = 10\sqrt 2 $$

$$4 \cdot \sqrt {7\frac{1}{2}} – \frac{{2\sqrt {10} }}{{2\sqrt 3 – \sqrt {10} }} = \sqrt {\frac{{16 \cdot 15}}{2}} – \frac{{2\sqrt {10} }}{{\sqrt 2 \left( {\sqrt 6 – \sqrt 5 } \right)}} = $$ $$\sqrt {120} – \frac{{2\sqrt 5 (\sqrt 6 + \sqrt 5 )}}{{\left( {\sqrt 6 – \sqrt 5 } \right)\left( {\sqrt 6 + \sqrt 5 } \right)}} = 2\sqrt {30} – 2\sqrt 5 \left( {\sqrt 6 + \sqrt 5 } \right) = $$ $$ = 2\sqrt {30} – 2\sqrt {30} – 10 = – 10$$

$$\frac{{\left( {\sqrt {75} + \sqrt {50} } \right)\left( {5 – 2\sqrt 6 } \right)}}{{\sqrt 3 – \sqrt 2 }} = \frac{{5\left( {\sqrt 3 + \sqrt 2 } \right)\left( {5 – 2\sqrt 6 } \right)}}{{\sqrt 3 – \sqrt 2 }} = $$ $$ = \frac{{5\left( {\sqrt 3 + \sqrt 2 } \right){{\left( {\sqrt 3 – \sqrt 2 } \right)}^2}}}{{\sqrt 3 – \sqrt 2 }} = 5\left( {\sqrt 3 + \sqrt 2 } \right)\left( {\sqrt 3 – \sqrt 2 } \right) = 5$$

$$\sqrt {\frac{2}{5}} + \sqrt {\frac{5}{2}} + \sqrt {10} = \frac{{\sqrt 2 }}{{\sqrt 5 }} + \frac{{\sqrt 5 }}{{\sqrt 2 }} + \sqrt {10} = $$ $$ = \frac{{\sqrt {10} }}{5} + \frac{{\sqrt {10} }}{2} + \sqrt {10} = \frac{{17\sqrt {10} }}{{10}} = 1,7\sqrt {10} $$

$$\left( {6 \cdot \sqrt[3]{{500}} – 5 \cdot \sqrt[3]{{108}}} \right):\sqrt[5]{{8 \cdot \sqrt[3]{2}}} = \left( {6 \cdot \sqrt[3]{{125 \cdot 4}} – 5 \cdot \sqrt[3]{{27 \cdot 4}}} \right):\sqrt[5]{{\sqrt[3]{{{2^9} \cdot 2}}}} = $$ $$ = \left( {30 \cdot \sqrt[3]{4} – 15 \cdot \sqrt[3]{4}} \right):\sqrt[{15}]{{{2^{10}}}} = 15 \cdot \sqrt[3]{4}:\sqrt[3]{4} = 15$$

$$3\sqrt {15\sqrt {75} } – 7\sqrt {6\sqrt {12} } = 3\sqrt {15\sqrt {25 \cdot 3} } – 7\sqrt {6\sqrt {4 \cdot 3} } = $$ $$ = 3\sqrt {75\sqrt 3 } – 7\sqrt {12\sqrt 3 } = 15\sqrt {3\sqrt 3 } – 14\sqrt {3\sqrt 3 } = \sqrt {3\sqrt 3 } = \sqrt[4]{{27}}$$

$$\sqrt[3]{{54 \cdot 32}} – \sqrt[4]{{8 \cdot 162}} + \sqrt[3]{{42\frac{7}{8}}} = \sqrt[3]{{27 \cdot 2 \cdot 32}} – $$ $$ – \sqrt[4]{{8 \cdot 81 \cdot 2}} + \sqrt[3]{{\frac{{343}}{8}}} = 3 \cdot 4 – 3 \cdot 2 + \frac{7}{2} = 9,5$$

$$\sqrt {\sqrt {55} \cdot \sqrt {275} \cdot \sqrt {605} } = \sqrt {\sqrt {55 \cdot 55 \cdot 5 \cdot 55 \cdot 11} } = \sqrt {55 \cdot 55} = 55$$