Есептеңіз.
№ 1 Есептеңіз: $(4\sqrt 6 + \sqrt {39} + 2\sqrt {26} + 6) \cdot (4\sqrt 6 + \sqrt {39} - 2\sqrt {26} - 6)$
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: $-5$
№ 2 Есептеңіз: $\left( {\sqrt {28} - \sqrt {12} } \right) \cdot \sqrt {10 + \sqrt {84} } $
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: $8$
№ 3 Есептеңіз: $\dfrac{{\sqrt {97 - 56\sqrt 3 } + 4\sqrt 3 }}{{\sqrt {52 - 30\sqrt 3 } - 3\sqrt 3 }} \cdot (\sqrt {3 + 2\sqrt 2 } + \sqrt {3 - 2\sqrt 2 } )$
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}$
№ 4 Есептеңіз: ${\left( {15 \cdot {4^{ - 2}} + \dfrac{{{2^{ - 4}} \cdot \sqrt {11} }}{{{{121}^{0,25}}}}} \right) \cdot \left( {\dfrac{{{{\left. {\left( {1 + {9^{0,25}}} \right) \cdot (\sqrt 3 - 1)} \right)}^{ - 1}}}}{4}} \right)}$
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: $2$
№ 5 Есептеңіз: ${(2\sqrt 6 - 5)^2} - 10\sqrt {49 - 20\sqrt 6 } + 1$
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: $0$
№ 6 Есептеңіз: $\sqrt 6 + \sqrt 5 - \dfrac{1}{{\sqrt {11 - 2\sqrt {30} } }} $
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: $0$
№ 7 Есептеңіз: $\sqrt[3]{{5\sqrt 2 - 7}} \cdot \sqrt {3 + 2\sqrt 2 } $
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: $1$
№ 8 Есептеңіз: $\dfrac{{\sqrt {21 + 8\sqrt 5 } }}{{4 + \sqrt 5 }} \cdot \sqrt {9 - 4\sqrt 5 } $
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: $\sqrt 5 - 2$
№ 9 Есептеңіз: $\dfrac{{\sqrt {6,3 \cdot 1,7} \cdot \left( {\sqrt {\dfrac{{6,3}}{{1,7}}} - \sqrt {\dfrac{{1,7}}{{6,3}}} } \right)}}{{\sqrt {{{(6,3 + 1,7)}^2} - 4 \cdot 6,3 \cdot 1,7} }}$
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: $1$
№ 10 Есептеңіз: $\left( {\dfrac{{\sqrt {{{561}^2} - {{459}^2}} }}{{4\dfrac{2}{7} \cdot 0,15 + 4\dfrac{2}{7}:\dfrac{{20}}{3}}} + 4\sqrt {10} } \right):\dfrac{{\sqrt {40} }}{3}$
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: $125$
№ 11 Есептеңіз: ${\left( {\dfrac{2}{{\sqrt 7 + \sqrt 5 }} + \dfrac{3}{{\sqrt 5 - \sqrt 2 }}} \right)^{ - 2}}:{\left( {1 + \sqrt {3,5} } \right)^{ - 2}} $
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: $\dfrac{1}{2}$
№ 12 Есептеңіз: ${\left( {\sqrt[6]{{9 + 4\sqrt 5 }} + \sqrt[3]{{2 + \sqrt 5 }}} \right) \cdot \sqrt[3]{{\sqrt 5 - 2}} }$
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: $2$
№ 13 Есептеңіз: ${{{\left( {1 + {2^{0,5}}} \right)}^2} + {{\left( {3 + 2\sqrt 2 } \right)}^{ - 1}}}$
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: $6$
№ 14 Есептеңіз: $\left( {{{\left( {\sqrt[4]{3} - \sqrt[4]{{27}}} \right)}^2} + 7} \right) \cdot \left( {{{\left( {\sqrt[4]{3} + \sqrt[4]{{27}}} \right)}^2} - 7} \right)$
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: $47$
№ 15 Есептеңіз: ${13 + 30\sqrt {2 + \sqrt {9 + 4\sqrt 2 } } }$
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: $5 + 3\sqrt 2$
№ 16 Есептеңіз: $\sqrt[3]{{\sqrt 5 + \sqrt[4]{{52}}}} \cdot \sqrt[3]{{5 + 2\sqrt {13} }} \cdot \sqrt[3]{{\sqrt[4]{{52}} - \sqrt 5 }} $
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: $3$
№ 17 Есептеңіз: $\dfrac{{\sqrt {12 - 2\sqrt {11} } }}{{\sqrt {17 - 12\sqrt 2 } }} - \dfrac{{\sqrt {12 + 2\sqrt {11} } }}{{\sqrt {17 + 12\sqrt 2 } }}$
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: $4\sqrt {22} - 6$
№ 18 Есептеңіз: ${7521^2} - 7522 \cdot 7520$
Шешуі: $$ = {7521^2} - \left( {7521 + 1} \right)\left( {7521 - 1} \right) = $$ $$ = {7521^2} - \left( {{{7521}^2} - {1^2}} \right) = 1$$ Жауабы: $1$
№ 19 Есептеңіз: $\sqrt[4]{{{{(\sqrt 5 - 3)}^4}}} - 3\sqrt 5 + 7$
Шешуі: $$ = \left| {\sqrt 5 - 3} \right| - 3\sqrt 5 + 7 = 3 - \sqrt 5 - 3\sqrt 5 + 7 = 10 - 4\sqrt 5 $$ Жауабы: $10 - 4\sqrt 5$
№ 20 Есептеңіз: $\dfrac{1}{{2 \cdot 5}} + \dfrac{1}{{5 \cdot 8}} + \dfrac{1}{{8 \cdot 11}} + \dfrac{1}{{11 \cdot 14}} + \dfrac{1}{{14 \cdot 17}}$
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: $\dfrac{5}{34}$
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