Сандық өрнектерді тепе-тең түрлендіру (Рустюмова 1.1.9)

Шешуі: $$ = \left( {\left( {4\sqrt 6 + \sqrt {39} } \right) + \left( {2\sqrt {26} + 6} \right)} \right) \cdot \left( {\left( {4\sqrt 6 + \sqrt {39} } \right) – \left( {2\sqrt {26} + 6} \right)} \right) = $$ $$ = {\left( {4\sqrt 6 + \sqrt {39} } \right)^2} – {\left( {2\sqrt {26} + 6} \right)^2} = $$ $$ = 96 + 39 + 8\sqrt {6 \cdot 39} – \left( {4 \cdot 26 + 36 + 24\sqrt {26} } \right) = $$ $$ = 135 + 8\sqrt {26 \cdot 9} – 140 – 24\sqrt {26} = – 5 + 24\sqrt {26} – 24\sqrt {26} = – 5.$$ Жауабы: $-5$

Шешуі: $$ = \left( {2\sqrt 7 – 2\sqrt 3 } \right) \cdot \sqrt {{{\left( {\sqrt 3 + \sqrt 7 } \right)}^2}} = \left( {2\sqrt 7 – 2\sqrt 3 } \right) \cdot \left( {\sqrt 3 + \sqrt 7 } \right) = $$ $$ = 2 \cdot \left( {\sqrt 7 – \sqrt 3 } \right) \cdot \left( {\sqrt 7 + \sqrt 3 } \right) = 2 \cdot \left( {7 – 3} \right) = 8$$ Жауабы: $8$

Шешуі: $$ = \dfrac{{\sqrt {{{\left( {7 – \sqrt {48} } \right)}^2}} + 4\sqrt 3 }}{{\sqrt {{{\left( {5 – \sqrt {27} } \right)}^2} – 3\sqrt 3 } }} \cdot \left( {\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {1 – \sqrt 2 } \right)}^2}} } \right) = $$ $$ = \dfrac{{7 – 4\sqrt 3 + 4\sqrt 3 }}{{3\sqrt 3 – 5 – 3\sqrt 3 }} \cdot (1 + \sqrt 2 + \sqrt 2 – 1) = \dfrac{7}{{ – 5}} \cdot 2\sqrt 2 = – \dfrac{{14\sqrt 2 }}{5}$$ Жауабы: $ – \dfrac{{14\sqrt 2 }}{5}$

Шешуі: $${ = \left( {15 \cdot \dfrac{1}{{16}} + \dfrac{{\dfrac{1}{{16}} \cdot \sqrt {11} }}{{\sqrt[4]{{121}}}}} \right) \cdot {{\left( {\dfrac{{(1 + \sqrt[4]{9})(\sqrt 3 – 1)}}{4}} \right)}^{ – 1}} = }$$ $${ = \left( {\dfrac{{15}}{{16}} + \dfrac{{\sqrt {11} }}{{16\sqrt {11} }}} \right) \cdot {{\left( {\dfrac{{(1 + \sqrt 3 )(\sqrt 3 – 1)}}{4}} \right)}^{ – 1}} = \left( {\dfrac{{15}}{{16}} + \dfrac{1}{{16}}} \right) \cdot 2 = 2}$$ Жауабы: $2$

Шешуі: $$ = {(2\sqrt 6 – 5)^2} – 10\sqrt {{{(2\sqrt 6 – 5)}^2}} + 1 = 49 – 20\sqrt 6 – 10 \cdot \left| {2\sqrt 6 – 5} \right| + 1 = $$ $$ = 49 – 20\sqrt 6 – 10 \cdot (5 – 2\sqrt 6 ) + 1 = 50 – 20\sqrt 6 – 50 + 20\sqrt 6 = 0$$ Жауабы: $0$

Шешуі: $$= \sqrt 6 + \sqrt 5 – \dfrac{{\sqrt {11 + 2\sqrt {30} } }}{{\sqrt {121 – 120} }}=$$ $$ = \sqrt 6 + \sqrt 5 – \dfrac{{\sqrt {{{(\sqrt 6 + \sqrt 5 )}^2}} }}{1} = \sqrt 6 + \sqrt 5 – \left| {\sqrt 6 + \sqrt 5 } \right| = $$ $$ = \sqrt 6 + \sqrt 5 – (\sqrt 6 + \sqrt 5 ) = \sqrt 6 + \sqrt 5 – \sqrt 6 – \sqrt 5 = 0$$ Жауабы: $0$

Шешуі: $$ = \sqrt[6]{{{{(5\sqrt 2 – 7)}^2}}} \cdot \sqrt[6]{{{{(3 + 2\sqrt 2 )}^3}}} = $$ $$ = \sqrt[6]{{50 – 14 \cdot 5\sqrt 2 + 49}} \cdot \sqrt[6]{{27 + 3 \cdot 18\sqrt 2 + 72 + 16\sqrt 2 }} = $$ $$ = \sqrt[6]{{9801 – 9800}} = 1.$$ Жауабы: $1$

Шешуі: $$= \dfrac{{\sqrt {16 + 2 \cdot 4 \cdot \sqrt 5 + 5} }}{{\sqrt 5 + 4}} \cdot \sqrt {{{(2 – \sqrt 5 )}^2}} =$$ $${ = \dfrac{{\sqrt {{{(4 + \sqrt 5 )}^2}} }}{{4 + \sqrt 5 }} \cdot \left| {2 – \sqrt 5 } \right| = \dfrac{{\left| {4 + \sqrt 5 } \right|}}{{4 + \sqrt 5 }} \cdot (\sqrt 5 – 2) = }$$ $${ = \dfrac{{4 + \sqrt 5 }}{{4 + \sqrt 5 }}(\sqrt 5 – 2) = \sqrt 5 – 2.}$$ Жауабы: $\sqrt 5 – 2$

Шешуі: $$ = \dfrac{{\sqrt {6,3 \cdot 1,7} \cdot \dfrac{{6,3 – 1,7}}{{\sqrt {6,3 \cdot 1,7} }}}}{{\sqrt {{{6,3}^2} + 2 \cdot 6,3 \cdot 1,7 + {{1,7}^2} – 4 \cdot 6,3 \cdot 1,7} }} = $$ $$ = \dfrac{{6,3 – 1,7}}{{\sqrt {{{\left( {6,3 – 1,7} \right)}^2}} }} = \dfrac{{4,6}}{{4,6}} = 1$$ Жауабы: $1$

Шешуі: $$ = \left( {\dfrac{{\sqrt {102 \cdot 1020} }}{{\dfrac{{30}}{7}\left( {0,15 + \dfrac{3}{{20}}} \right)}} + 4\sqrt {10} } \right) \cdot \dfrac{3}{{\sqrt {40} }} = $$ $$=\left( { \dfrac{{2 \cdot 7 \cdot 17\sqrt {10} + 12\sqrt {10} }}{3}} \right) \cdot \dfrac{3}{{\sqrt {40} }} = \dfrac{{250\sqrt {10} }}{{\sqrt {40} }} = 125$$ Жауабы: $125$

Шешуі: $$ = {\left( {\dfrac{{2(\sqrt 7 – \sqrt 5 )}}{2} + \dfrac{{3(\sqrt 5 + \sqrt 2 )}}{3}} \right)^{ – 2}}:{\left( {1 + \sqrt {\dfrac{7}{2}} } \right)^{ – 2}} = $$ $$ = {\left( {\sqrt 7 – \sqrt 5 + \sqrt 5 + \sqrt 2 } \right)^{ – 2}}:{\left( {\dfrac{{\sqrt 2 + \sqrt 7 }}{{\sqrt 2 }}} \right)^{ – 2}} = $$ $$ = {\left( {\sqrt 7 + \sqrt 2 } \right)^{ – 2}}:{\left( {\dfrac{{\sqrt 7 + \sqrt 2 }}{{\sqrt 2 }}} \right)^{ – 2}} = {\left( {\dfrac{{\left( {\sqrt 7 + \sqrt 2 } \right) \cdot \sqrt 2 }}{{\left( {\sqrt 7 + \sqrt 2 } \right)}}} \right)^{ – 2}} = $$ $$ = {\left( {\sqrt 2 } \right)^{ – 2}} = {\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2} = \dfrac{1}{2}$$ Жауабы: $\dfrac{1}{2}$

Шешуі: $${ = \left( {\sqrt[6]{{{{\left( {2 + \sqrt 5 } \right)}^2}}} + \sqrt[3]{{2 + \sqrt 5 }}} \right) \cdot \sqrt[3]{{\sqrt 5 – 2}} = }$$ $${ = \left( {\sqrt[3]{{2 + \sqrt 5 }} + \sqrt[3]{{2 + \sqrt 5 }}} \right) \cdot \sqrt[3]{{\sqrt 5 – 2}} = }$$ $${ = 2\sqrt[3]{{2 + \sqrt 5 }} \cdot \sqrt[3]{{\sqrt 5 – 2}} = 2\sqrt {5 – 4} = 2 \cdot 1 = 2.}$$ Жауабы: $2$

Шешуі: $${ = {{\left( {1 + \sqrt 2 } \right)}^2} + {{\left( {3 + 2\sqrt 2 } \right)}^{ – 1}} = }$$ $${1 + 2\sqrt 2 + 2 + \dfrac{1}{{3 + 2\sqrt 2 }} = 3 + 2\sqrt 2 + \dfrac{{3 – 2\sqrt 2 }}{{9 – 8}} = }$$ $${ = 3 + 2\sqrt 2 + 3 – 2\sqrt 2 = 6}$$ Жауабы: $6$

Шешуі: $$ = \left( {\sqrt 3 – 2\sqrt[4]{{81}} + \sqrt {27} + 7} \right) \cdot \left( {\sqrt 3 + 2\sqrt[4]{{81}} + \sqrt {27} – 7} \right) = $$ $$ = (4\sqrt 3 + 1) \cdot (4\sqrt 3 – 1) = 48 – 1 = 47$$ Жауабы: $47$

Шешуі: $$ = \sqrt {13 + 30\sqrt {2 + \sqrt {{{\left( {1 + \sqrt 8 } \right)}^2}} } } = \sqrt {13 + 30\sqrt {2 + 1 + \sqrt 8 } } = $$ $$ = \sqrt {13 + 30\sqrt {2 + 1 + 2\sqrt 2 } } = \sqrt {13 + 30\sqrt {{{\left( {\sqrt 2 } \right)}^2} + 2 \cdot \sqrt 2 \cdot 1 + {1^2}} } = $$ $$ = \sqrt {13 + 30\sqrt {{{\left( {1 + \sqrt 2 } \right)}^2}} } = \sqrt {13 + 30 \cdot \left( {1 + \sqrt 2 } \right)} = $$ $$ = \sqrt {13 + 30 + 30\sqrt 2 } = \sqrt {43 + 30\sqrt 2 } = $$ $$ = \sqrt {{{(5 + 3\sqrt 2 )}^2}} = 5 + 3\sqrt 2 $$ Жауабы: $5 + 3\sqrt 2$

Шешуі: $${ = \sqrt[3]{{(\sqrt[4]{{52}} + \sqrt 5 )(\sqrt[4]{{52}} – \sqrt 5 )(2\sqrt {13} + 5)}} = }$$ $${ = \sqrt[3]{{(\sqrt {52} – 5)(\sqrt {52} + 5)}} = \sqrt[3]{{52 – 25}} = \sqrt[3]{{27}} = 3}$$ Жауабы: $3$

Шешуі: $$ = \dfrac{{\sqrt {{{(1 – \sqrt {11} )}^2}} }}{{\sqrt {{{(3 – 2\sqrt 2 )}^2}} }} – \dfrac{{\sqrt {{{(1 + \sqrt {11} )}^2}} }}{{\sqrt {{{(3 + 2\sqrt 2 )}^2}} }} = \dfrac{{\sqrt {11} – 1}}{{3 – 2\sqrt 2 }} – \dfrac{{1 + \sqrt {11} }}{{3 + 2\sqrt 2 }} = $$ $$ = (\sqrt {11} – 1)(3 + 2\sqrt 2 ) – (1 + \sqrt {11} )(3 – 2\sqrt 2 ) = $$ $$ = 3\sqrt {11} – 3 + 2\sqrt {22} – 2\sqrt 2 – 3 – 3\sqrt {11} + 2\sqrt 2 + 2\sqrt {22} = 4\sqrt {22} – 6$$ Жауабы: $4\sqrt {22} – 6$

Шешуі: $$ = {7521^2} – \left( {7521 + 1} \right)\left( {7521 – 1} \right) = $$ $$ = {7521^2} – \left( {{{7521}^2} – {1^2}} \right) = 1$$ Жауабы: $1$

Шешуі: $$ = \left| {\sqrt 5 – 3} \right| – 3\sqrt 5 + 7 = 3 – \sqrt 5 – 3\sqrt 5 + 7 = 10 – 4\sqrt 5 $$ Жауабы: $10 – 4\sqrt 5$

Шешуі: $${ = \left( {\dfrac{1}{2} – \dfrac{2}{5}} \right) + \left( {\dfrac{2}{5} – \dfrac{3}{8}} \right) + \left( {\dfrac{3}{8} – \dfrac{4}{{11}}} \right) + \left( {\dfrac{4}{{11}} – \dfrac{5}{{14}}} \right) + \left( {\dfrac{5}{{14}} – \dfrac{6}{{17}}} \right) = }$$ $${ = \dfrac{1}{2} – \dfrac{2}{5} + \dfrac{2}{5} – \dfrac{3}{8} + \dfrac{3}{8} – \dfrac{4}{{11}} + \dfrac{4}{{11}} – \dfrac{5}{{14}} + \dfrac{5}{{14}} – \dfrac{6}{{17}} = }$$ $${ = \dfrac{1}{2} – \dfrac{6}{{17}} = \dfrac{5}{{34}}}$$ Жауабы: $\dfrac{5}{34}$