Рустюмова 1.1.3

 +/-  - Есептің жауабын көрсету/көрсетпеу.

▲/▼ - Жауап орнын жасыру/шығару

   ×    - Сұрақты алып тастау.

№ 1.1 Есептеңіз: ${\dfrac{{{7^8} \cdot {{49}^{ - 2}} \cdot {5^4} + 49 \cdot 125 \cdot {{\left( {\dfrac{1}{5}} \right)}^{ - 1}}}}{{{{(7 \cdot 5)}^4} \cdot {7^{ - 3}}}}}$

Шешуі: $${\dfrac{{{7^8} \cdot {{\left( {{7^2}} \right)}^{ - 2}} \cdot {5^4} + {7^2} \cdot {5^3} \cdot {{\left( {{5^{ - 1}}} \right)}^{ - 1}}}}{{{7^4} \cdot {5^4} \cdot {7^{ - 3}}}} = \dfrac{{{7^8} \cdot {7^{ - 4}} \cdot {5^4} + {7^2} \cdot {5^3} \cdot 5}}{{{7^{4 - 3}} \cdot {5^4}}} = }$$ $${ = \dfrac{{{7^4} \cdot {5^4} + {7^2} \cdot {5^4}}}{{7 \cdot {5^4}}} = \dfrac{{{5^4} \cdot {7^2}\left( {{7^2} + 1} \right)}}{{7 \cdot {5^4}}} = \dfrac{{7 \cdot 50}}{1} = 350}$$

№ 1.2 Есептеңіз: ${\dfrac{{\left( {3 \cdot {2^{20}} + 7 \cdot {2^{19}}} \right) \cdot 52}}{{{{\left( {13 \cdot {8^4}} \right)}^2}}}}$

Шешуі: $${ = \dfrac{{{2^{19}}(3 \cdot 2 + 7) \cdot 52}}{{{{13}^2} \cdot {{\left( {{2^3}} \right)}^8}}} = \dfrac{{{2^{19}} \cdot 13 \cdot (13 \cdot 4)}}{{{{13}^2} \cdot {2^{24}}}} = }$$ $${ = \dfrac{{{2^{19}} \cdot {{13}^2} \cdot {2^2}}}{{{{13}^2} \cdot {2^{24}}}} = \dfrac{{{2^{19 + 2 - 24}}}}{1} = {2^{ - 3}} = \dfrac{1}{{{2^3}}} = \dfrac{1}{8}}$$

№ 1.3 Есептеңіз: $\dfrac{{\root 3 \of 3 \cdot \root 4 \of 9 \cdot \root 6 \of {81} }}{{\sqrt 3 }}$

Шешуі: $$ = \frac{{{3^{\frac{1}{3}}} \cdot {{\left( {{3^2}} \right)}^{\frac{1}{4}}} \cdot {{\left( {{3^4}} \right)}^{\frac{1}{6}}}}}{{{3^{\frac{1}{2}}}}} = \frac{{{3^{\frac{1}{3}}} \cdot {3^{\frac{1}{2}}} \cdot {3^{\frac{2}{3}}}}}{{{3^{\frac{1}{2}}}}} = {3^{\frac{1}{3} + \frac{2}{3}}} = 3$$

№ 1.4 Есептеңіз: $\dfrac{{\root 8 \of {27} \cdot \root 3 \of {40} \cdot \root 4 \of 4 \cdot \root 8 \of {{3^5}} }}{{\root 6 \of {25} \cdot \sqrt 2 }}$

Шешуі: $$ = \frac{{\root 8 \of {{3^3}} \cdot \root 3 \of {8 \cdot 5} \cdot \root 4 \of {{2^2}} \cdot \root 8 \of {{3^5}} }}{{\root 6 \of {{5^2}} \cdot \sqrt 2 }} = \frac{{{3^{\frac{3}{8}}} \cdot 2 \cdot {5^{\frac{1}{3}}} \cdot {2^{\frac{1}{2}}} \cdot {3^{\frac{5}{8}}}}}{{{5^{\frac{1}{3}}} \cdot {2^{\frac{1}{2}}}}} = $$ $$ = \frac{{{3^{\frac{3}{8} + \frac{5}{8}}} \cdot 2}}{1} = \frac{{3 \cdot 2}}{1} = 6$$

№ 1.5 Есептеңіз: $\dfrac{{(8 + \sqrt {56} )(1,5 + 0,25)}}{{\sqrt 2 \cdot (\sqrt 7 + \sqrt 8 )}}$

Шешуі: $$ = \dfrac{{(8 + \sqrt {4 \cdot 14} ) \cdot 1,75}}{{\sqrt 2 \cdot (\sqrt 7 + \sqrt {4 \cdot 2} )}} = \dfrac{{(14 + 3,5 \cdot \sqrt {14} )(\sqrt {14} - 4)}}{{(\sqrt {14} + 4)(\sqrt {14} - 4)}} = $$ $$ = \dfrac{{14\sqrt {14} - 56 + 49 - 14\sqrt {14} }}{{14 - 16}} = \dfrac{{ - 7}}{{ - 2}} = \dfrac{7}{2} = 3,5$$

№ 1.6 Есептеңіз: ${\dfrac{{(\sqrt {45} - \sqrt {20} )(\sqrt {12} + \sqrt {75} ) \cdot 7\sqrt 3 }}{{\sqrt 5 + \sqrt {180} }}}$

Шешуі: $${ = \dfrac{{(\sqrt {9 \cdot 5} - \sqrt {4 \cdot 5} ) \cdot (\sqrt {4 \cdot 3} + \sqrt {25 \cdot 3} ) \cdot 7\sqrt 3 }}{{\sqrt 5 + \sqrt {36 \cdot 5} }} = }$$ $${ = \dfrac{{(3\sqrt 5 - 2\sqrt 5 ) \cdot (2\sqrt 3 + 5\sqrt 3 ) \cdot 7\sqrt 3 }}{{\sqrt 5 + 6\sqrt 5 }} = \dfrac{{\sqrt 5 \cdot 7\sqrt 3 \cdot 7\sqrt 3 }}{{7\sqrt 5 }} = }$$ $${ = 7 \cdot 3 = 21}$$

№ 1.7 Есептеңіз: ${\dfrac{{{{(2 - \sqrt 5 )}^2}}}{{(3 - 2 \cdot \root 4 \of 5 )(3 + 2 \cdot \root 4 \of 5 )}}}$

Шешуі: $${ = \dfrac{{4 - 4\sqrt 5 + 5}}{{{3^2} - {{(2\root 4 \of 5 )}^2}}} = \dfrac{{9 - 4\sqrt 5 }}{{9 - \sqrt 5 }} = 1}$$

№ 1.8 Есептеңіз: ${\dfrac{{1 - 2 \cdot \root 4 \of 5 + \sqrt 5 }}{{{{(\sqrt 3 - \root 4 \of {45} )}^2}}}}$

Шешуі: $${ = \dfrac{{1 - 2 \cdot \root 4 \of 5 + \sqrt 5 }}{{3 - 2 \cdot \sqrt 3 \cdot \root 4 \of {{3^2} \cdot 5} + \sqrt {{9^2} \cdot 5} }} = \dfrac{{1 - 2 \cdot \root 4 \of 5 + \sqrt 5 }}{{3 - 2 \cdot 3 \cdot \root 4 \of 5 + 3\sqrt 5 }} = }$$ $${\dfrac{{1 - 2\root 4 \of 5 + \sqrt 5 }}{{3(1 - 2\root 4 \of 5 + \sqrt 5 )}} = \dfrac{1}{3}}$$

№ 1.9 Есептеңіз: $$\dfrac{{(\root 3 \of 5 - \root 3 \of 9 ){{(\sqrt {32} - \sqrt 8 )}^2}}}{{\root 3 \of {40} - \root 3 \of {72} }}$$

Шешуі: $$ = \dfrac{{(\root 3 \of 5 - \root 3 \of 9 ){{\left( {\sqrt {{2^5}} - \sqrt {{2^3}} } \right)}^2}}}{{\root 3 \of {{2^3} \cdot 5} - \root 3 \of {9 \cdot {2^3}} }} = \dfrac{{(\root 3 \of 5 - \root 3 \of 9 ) \cdot \left( {32 - 2\sqrt {{2^8}} + 8} \right)}}{{2\root 3 \of 5 - 2\root 3 \of 9 }} = $$ $$ = \dfrac{{(\root 3 \of 5 - \root 3 \of 9 )\left( {40 - 2 \cdot {2^4}} \right)}}{{2(\root 3 \of 5 - \root 3 \of 9 )}} = \dfrac{{40 - 32}}{2} = 4$$

№ 1.10 Есептеңіз: $\dfrac{{(\sqrt 3 - \sqrt 2 ) \cdot \sqrt {72} }}{{3(2\sqrt 6 - \sqrt {16} )(\root 3 \of {64} + 1)}}$

Шешуі: $$ = \dfrac{{(\sqrt 3 - \sqrt 2 ) \cdot \sqrt {36 \cdot 2} }}{{3 \cdot (2\sqrt 6 - 4) \cdot \left( {\root 3 \of {{2^6}} + 1} \right)}} = \dfrac{{(\sqrt 3 - \sqrt 2 ) \cdot 6\sqrt 2 }}{{3 \cdot 2(\sqrt 6 - 2) \cdot (4 + 1)}} = $$ $$ = \dfrac{{6\sqrt 6 - 12}}{{30(\sqrt 6 - 2)}} = \dfrac{{6(\sqrt 6 - 2)}}{{30(\sqrt 6 - 2)}} = \dfrac{1}{5} = 0,2$$

№ 1.11 Есептеңіз: $\dfrac{{\sqrt 6 + \sqrt 3 - \sqrt 2 - 1}}{{\sqrt 6 + 2\sqrt 3 - \sqrt 2 - 2}}$

Шешуі: $$ = \dfrac{{\sqrt 3 (\sqrt 2 + 1) - (\sqrt 2 + 1)}}{{\sqrt 3 (\sqrt 2 + 2) - (\sqrt 2 + 2)}} = \dfrac{{(\sqrt 2 + 1)(\sqrt 3 - 1)}}{{(\sqrt 2 + 2)(\sqrt 3 - 1)}} = $$ $$ = \dfrac{{(\sqrt 2 + 1)(\sqrt 2 - 2)}}{{(\sqrt 2 + 2)(\sqrt 2 - 2)}} = \dfrac{{2 - 2\sqrt 2 + \sqrt 2 - 2}}{{2 - 4}} = \dfrac{{ - \sqrt 2 }}{{ - 2}} = \dfrac{{\sqrt 2 }}{2}$$

№ 1.12 Есептеңіз: $\dfrac{{(\sqrt 5 - \sqrt {11} )(\sqrt {33} + \sqrt {15} - \sqrt {22} - \sqrt {10} )}}{{\sqrt {75} - \sqrt {50} }}$

Шешуі: $$ = \dfrac{{(\sqrt 5 - \sqrt {11} ) \cdot (\sqrt {3 \cdot 11} + \sqrt {5 \cdot 3} - \sqrt {2 \cdot 11} - \sqrt {5 \cdot 2} )}}{{\sqrt {25 \cdot 3} - \sqrt {25 \cdot 2} }} = $$ $$ = \dfrac{{(\sqrt 5 - \sqrt {11} ) \cdot \left( {\sqrt 3 (\sqrt {11} + \sqrt 5 ) - \sqrt 2 (\sqrt {11} + \sqrt 5 )} \right)}}{{5\sqrt 3 - 5\sqrt 2 }} = $$ $$ = \dfrac{{(\sqrt 5 - \sqrt {11} )(\sqrt 3 - \sqrt 2 )(\sqrt {11} + \sqrt 5 )}}{{5(\sqrt 3 - \sqrt 2 )}} = \dfrac{{5 - 11}}{5} = \dfrac{{ - 6}}{5} = - 1,2$$

№ 1.13 Есептеңіз: $\dfrac{{{{(\root 3 \of 9 + \sqrt 3 )}^2}}}{{\root 3 \of 3 + 2 \cdot \root 6 \of 3 + 1}}$

Шешуі: $$ = \dfrac{{\root 3 \of {81} + 2 \cdot \root 3 \of 9 \cdot \sqrt 3 + 3}}{{\root 3 \of 3 + 2\root 6 \of 3 + 1}} = \dfrac{{3\root 3 \of 3 + 2\root 6 \of {{9^2}} \cdot \root 6 \of {{3^3}} + 3}}{{\root 3 \of 3 + 1 + 2\root 6 \of 3 }} = $$ $$ = \dfrac{{3\root 3 \of 3 + 2 \cdot \root 6 \of {{3^7}} + 3}}{{\root 3 \of 3 + 1 + 2\root 6 \of 3 }} = \dfrac{{3\root 3 \of 3 + 6\root 6 \of 3 + 3}}{{\root 3 \of 3 + 1 + 2\root 6 \of 3 }} = $$ $$ = \dfrac{{(3\root 3 \of 3 + 6\root 6 \of 3 + 3)(\root 3 \of 3 + 1 - 2\root 6 \of 3 )}}{{(\root 3 \of 3 + 1 + 2\root 6 \of 3 )(\root 3 \of 3 + 1 - 2\root 6 \of 3 )}} = $$ $$ = \dfrac{{3\root 3 \of 9 + 3\root 3 \of 3 - 6\root 3 \of 3 \cdot \root 6 \of 3 + 6\root 6 \of 3 \cdot \root 3 \of 3 + 6\root 6 \of 3 - 12\root 6 \of 9 + 3\root 3 \of 3 + 3 - 6\root 6 \of 3 }}{{{{(\root 3 \of 3 + 1)}^2} - 4\root 3 \of 3 }} = $$ $$ = \dfrac{{3\root 3 \of 9 + 3\root 3 \of 3 - 12\root 6 \of {{3^2}} + 3\root 3 \of 3 + 3}}{{\root 3 \of 9 + 2\root 3 \of 3 + 1 - 4\root 3 \of 3 }} = $$ $$ = \dfrac{{3\root 3 \of 9 + 3\root 3 \of 3 - 12\root 3 \of 3 + 3\root 3 \of 3 + 3}}{{\root 3 \of 9 - 2\root 3 \of 3 + 1}} = $$ $${ = \dfrac{{3\sqrt[3]{9} - 6\sqrt[3]{3} + 3}}{{\sqrt[3]{9} - 2\sqrt[3]{3} + 1}} = \dfrac{{3(\sqrt[3]{9} - 2\sqrt[3]{3} + 1)}}{{\sqrt[3]{9} - 2\sqrt[3]{3} + 1}} = 3}$$

№ 1.14 Есептеңіз: ${\dfrac{25^{k+1}-25^k}{\left(5^k+5^{k-1}\right)^2}}$

Шешуі: $${ = \dfrac{{{{25}^k}(25 - 1)}}{{{5^{2k}} + 2 \cdot {5^k} \cdot {5^{k - 1}} + {5^{2k - 2}}}} = \dfrac{{{5^{2k}} \cdot 24}}{{{5^{2k}} + 2 \cdot {5^{2k - 1}} + {5^{2k - 2}}}} = }$$ $${ = \dfrac{{{5^{2k}} \cdot 24}}{{{5^{2k}}\left( {1 + 2 \cdot {5^{ - 1}} + {5^{ - 2}}} \right)}} = \dfrac{{24}}{{1 + \dfrac{2}{5} + \dfrac{1}{{25}}}} = \dfrac{{24}}{{\dfrac{{36}}{{25}}}} = }$$ $${ = 24 \cdot \dfrac{{25}}{{36}} = \dfrac{{50}}{3} = 16\dfrac{2}{3}}$$

№ 1.15 Есептеңіз: ${\dfrac{{{{25}^{k + 1}} - {{25}^k}}}{{{{\left( {{5^k} + {5^{k - 1}}} \right)}^2}}}}$

Шешуі: $${ = \dfrac{{{3^{2k}} + 2 \cdot {3^k} \cdot {3^{k - 1}} + {3^{2k - 2}}}}{{{3^{2k - 2}}}} = \dfrac{{{3^{2k}} + 2 \cdot {3^{2k}} \cdot {3^{ - 1}} + {3^{2k}} \cdot {3^{ - 2}}}}{{{3^{2k}} \cdot {3^{ - 2}}}} = }$$ $${ = \dfrac{{{3^{2k}}\left( {\dfrac{3}{1} + \dfrac{2}{3} + \dfrac{1}{9}} \right)}}{{{3^{2k}} \cdot \dfrac{1}{9}}} = \dfrac{{\dfrac{{16}}{9}}}{{\dfrac{1}{9}}} = \dfrac{{16}}{9} \cdot \dfrac{9}{1} = 16}$$

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