Рустюмова 1.1.2

 +/-  - Есептің жауабын көрсету/көрсетпеу.

▲/▼ - Жауап орнын жасыру/шығару

   ×    - Сұрақты алып тастау.

№ 1 Есептеңіз: $\dfrac{{{{41}^2} - {{17}^2}}}{{{{37}^2} - {{21}^2}}} - \dfrac{{{{39}^2} - {{27}^2}}}{{{{45}^2} - {{21}^2}}}$

Шешуі: $$ = \dfrac{{(41 - 17)(41 + 17)}}{{(37 - 21)(37 + 21)}} - \dfrac{{(39 - 27)(39 + 27)}}{{(45 - 21)(45 + 21)}} = $$ $$ = \dfrac{{24 \cdot 58}}{{16 \cdot 58}} - \dfrac{{12 \cdot 66}}{{24 \cdot 66}} = \dfrac{3}{2} - \dfrac{1}{2} = 1$$

№ 2 Есептеңіз: ${\left( {5\dfrac{1}{3} + \dfrac{3}{{16}}} \right)^2} - {\left( {5\dfrac{1}{3} - \dfrac{3}{{16}}} \right)^2} + {\left( {7\dfrac{3}{5} + \dfrac{5}{{38}}} \right)^2} - {\left( {7\dfrac{3}{5} - \dfrac{5}{{38}}} \right)^2}$

Шешуі: $$ = \left( {5\dfrac{1}{3} + \dfrac{3}{{16}} - 5\dfrac{1}{3} + \dfrac{3}{{16}}} \right)\left( {5\dfrac{1}{3} + \dfrac{3}{{16}} + 5\dfrac{1}{3} - \dfrac{3}{{16}}} \right) + \left( {7\dfrac{3}{5} + \dfrac{5}{{38}} - + \dfrac{3}{5} + \dfrac{5}{{38}}} \right)\left( {7\dfrac{3}{5} + \dfrac{5}{{38}} + 7\dfrac{3}{5} - \dfrac{5}{{38}}} \right) = $$ $$\dfrac{6}{{16}} \cdot 10\dfrac{2}{3} + \dfrac{{10}}{{38}} \cdot 14\dfrac{6}{5} = \dfrac{3}{8} \cdot \dfrac{{32}}{3} + \dfrac{5}{{19}} \cdot \dfrac{{76}}{5} = 4 + 4 = 8$$

№ 3 Есептеңіз: $\dfrac{{{{52}^2} - {{37}^2}}}{{{{57}^2} - {{32}^2}}} + \dfrac{{{{39}^2} - {{36}^2}}}{{{{45}^2} - {{30}^2}}}$

Шешуі: $${ = \dfrac{{(52 - 37)(52 + 37)}}{{(57 - 32)(57 + 32)}} + \dfrac{{(39 - 36)(39 + 36)}}{{(45 - 30)(45 + 30)}} = }$$ $${ = \dfrac{{15 \cdot 89}}{{25 \cdot 89}} + \dfrac{{3 \cdot 75}}{{15 \cdot 75}} = \dfrac{3}{5} + \dfrac{3}{{15}} = \dfrac{3}{5} + \dfrac{1}{5} = \dfrac{4}{5}}$$

№ 4 Есептеңіз: ${\dfrac{{{{7,4}^2} - {{2,6}^2}}}{{{{11,2}^2} - {{8,8}^2}}}}$

Шешуі: $${ = \dfrac{{(7,4 - 2,6)(7,4 + 2,6)}}{{(11,2 - 8,8)(11,2 + 8,8)}} = \dfrac{{4,8 \cdot 10}}{{2,4 \cdot 20}} = 1}$$

№ 5 Есептеңіз: $\dfrac{{{{3,88}^2} + {{3,12}^2}}}{2} + 3,88 \cdot 3,12$

Шешуі: $$ = \dfrac{{{{3,88}^2} + {{3,12}^2} + 2 \cdot 3,88 \cdot 3,12}}{2} = \dfrac{{{{(3,88 + 3,12)}^2}}}{2} = \dfrac{{49}}{2} = 24,5$$

№ 6 Есептеңіз: ${2,85^2} + {7,15^2} + \dfrac{{2,85 \cdot 7,15}}{{0,5}}$

Шешуі: $${2,85^2} + {7,15^2} + 2 \cdot 2,85 \cdot 7,15 = {(2,85 + 7,15)^2} = {10^2} = 100$$

№ 7 Есептеңіз: ${(\sqrt[3]{2} - \sqrt[3]{5})(\sqrt[3]{4} + \sqrt[3]{{10}} + \sqrt[3]{{25}})}$

Шешуі: $${(\sqrt[3]{2} - \sqrt[3]{5})\left( {{{(\sqrt[3]{2})}^2} + \sqrt[3]{5} \cdot \sqrt[3]{2} + {{(\sqrt[3]{5})}^2}} \right) = {{(\sqrt[3]{2})}^3} - {{(\sqrt[3]{5})}^3} = 2 - 5 = - 3}$$

№ 8 Есептеңіз: ${{{\left( {5 + {{17}^{\frac{1}{2}}}} \right)}^{ - \frac{1}{3}}}:{{(5 - \sqrt {17} )}^{\frac{1}{3}}}}$

Шешуі: $${ = \dfrac{1}{{\sqrt[3]{{5 + \sqrt {17} }}}} \cdot \dfrac{1}{{\sqrt[3]{{5 - \sqrt {17} }}}} = \dfrac{1}{{\sqrt[3]{{25 - 17}}}} = \dfrac{1}{{\sqrt[3]{8}}} = \dfrac{1}{2}}$$

№ 9 Есептеңіз: ${\sqrt {\dfrac{{\sqrt {{{56}^2} - {{46}^2}} }}{{0,25 \cdot \sqrt {10} }}} }$

Шешуі: $${ = \sqrt {\dfrac{{\sqrt {(56 - 46)(56 + 46)} }}{{\dfrac{1}{4} \cdot \sqrt {10} }}} = \sqrt {\dfrac{{4\sqrt {10} \cdot \sqrt {102} }}{{\sqrt {10} }}} = 2\sqrt[4]{{102}}}$$

№ 10 Есептеңіз: ${\dfrac{{7{{(\sqrt 5 - \sqrt 3 )}^2}}}{{20(2 + \sqrt[4]{{15}})(2 - \sqrt[4]{{15}})}}}$

Шешуі: $${ = \dfrac{{7{{(\sqrt 5 - \sqrt 3 )}^2}}}{{20(4 - \sqrt {15} )}} = \dfrac{{7(8 - 2\sqrt {15} )}}{{20(4 - \sqrt {15} )}} = \dfrac{{14(4 - \sqrt {15} )}}{{20(4 - \sqrt {15} )}} = 0,7}$$

№ 11 Есептеңіз: ${\sqrt[4]{{\sqrt {23} - \sqrt 7 }} \cdot \sqrt[4]{{\sqrt {23} + \sqrt 7 }}}$

Шешуі: $${ = \sqrt[4]{{{{(\sqrt {23} )}^2} - {{(\sqrt 7 )}^2}}} = \sqrt[4]{{23 - 7}} = \sqrt[4]{{16}} = 2}$$

№ 12 Есептеңіз: ${0,507^3} + {0,493^3} - 0,507 \cdot 0,493$

Шешуі: $${ = (0,507 + 0,493)\left( {{{0,507}^2} - 0,507 \cdot 0,493 + {{0,493}^2}} \right) - 0,507 \cdot 0,493 = }$$ $${ = {{0,507}^2} - 2 \cdot 0,507 \cdot 0,493 + {{0,493}^2} = }$$ $${ = {{(0,507 - 0,493)}^2} = {{0,014}^2} = 0,000196}$$

№ 13 Есептеңіз: ${\dfrac{{\dfrac{{{{57}^3} + {{63}^3}}}{{120}} - 57 \cdot 63}}{{{{18,5}^2} - {{17,5}^2}}}}$

Шешуі: $${ = \dfrac{{(57 + 63)\left( {{{57}^2} - 5 \cdot 63 + {{63}^2}} \right)}}{{120}} - 57 \cdot 63 = }$$ $${ = \dfrac{{(18,5 - 17,5)(18,5 + 17,5)}}{{1 \cdot 36}} = \dfrac{{36}}{{36}} = 1}$$

№ 14 Есептеңіз: $\dfrac{{\dfrac{{{{15,1}^3} - {{10,5}^3}}}{{4,6}} + 15,1 \cdot 10,5}}{{{{32}^2} - {{19,2}^2}}}$

Шешуі: $$ = \dfrac{{\dfrac{{(15,1 - 10,5)\left( {{{15,1}^2} + 15,1 \cdot 10,5 + {{10,5}^2}} \right)}}{{4,6}} + 15,1 \cdot 10,5}}{{{{32}^2} - {{19,2}^2}}} = $$ $$ = \dfrac{{{{(15,1 + 10,5)}^2}}}{{(32 - 19,2)(32 + 19,2)}} = \dfrac{{{{25,6}^2}}}{{12,8 \cdot 51,2}} = \dfrac{{25,6 \cdot 25,6}}{{12,8 \cdot 25,6 \cdot 2}} = 1$$

№ 15 Есептеңіз: ${\left( {{9^{ - 0,25}} + {{(2\sqrt 2 )}^{ - \frac{2}{3}}}} \right)\left( {{9^{ - 0,25}} - {{(2\sqrt 2 )}^{ - \frac{2}{3}}}} \right)}$

Шешуі: $${{9^{ - 0,25}} = {{\left( {{3^2}} \right)}^{ - 0,25}} = {3^{ - 0,5}} = {3^{ - \frac{1}{2}}} = \frac{1}{{\sqrt 3 }}}$$ $${{{(2\sqrt 2 )}^{ - \frac{2}{3}}} = {{\left( {\sqrt {{2^3}} } \right)}^{ - \frac{2}{3}}} = {2^{\frac{3}{2} \cdot \left( { - \frac{2}{3}} \right)}} = {2^{ - 1}} = \frac{1}{2}}$$ $${\left( {\frac{1}{{\sqrt 3 }} + \frac{1}{2}} \right)\left( {\frac{1}{{\sqrt 3 }} - \frac{1}{2}} \right) = \frac{1}{3} - \frac{1}{4} = \frac{1}{{12}}}$$

№ 16 Есептеңіз: $\sqrt {\dfrac{{{{(\sqrt {13} )}^3} + {{(\sqrt 3 )}^3}}}{{\sqrt {13} + \sqrt 3 }} - \sqrt {39} } \cdot (\sqrt {13} + \sqrt 3 )$

Шешуі: $$ = \sqrt {\dfrac{{(\sqrt {13} + \sqrt 3 )\left( {{{(\sqrt {13} )}^2} - \sqrt {39} + {{(\sqrt 3 )}^2}} \right)}}{{\sqrt {13} + \sqrt 3 }} - \sqrt {39} } \cdot (\sqrt {13} + \sqrt 3 ) = $$ $$ = \sqrt {{{(\sqrt {13} - \sqrt 3 )}^2}} - (\sqrt {13} + \sqrt 3 ) = (\sqrt {13} - \sqrt 3 )(\sqrt {13} + \sqrt 3 ) = 13 - 3 = 10$$

№ 17 Есептеңіз: ${\dfrac{{6{{(\sqrt 2 - 1)}^3}}}{{1 - 5{{(\sqrt 2 - 1)}^2}}}}$

Шешуі: $${ = \dfrac{{6\left( {{{(\sqrt 2 )}^3} - 3{{(\sqrt 2 )}^2} \cdot 1 + 3 \cdot \sqrt 2 \cdot {1^2} - {1^3}} \right)}}{{1 - 5\left( {{{(\sqrt 2 )}^2} - 2\sqrt 2 \cdot 1 + {1^2}} \right)}} = }$$ $${ = \dfrac{{6(2\sqrt 2 - 6 + 3\sqrt 2 - 1)}}{{1 - 5(3 - 2\sqrt 2 )}} = \dfrac{{6(5\sqrt 2 - 7)}}{{1 - 15 + 10\sqrt 2 }} = \dfrac{{6(5\sqrt 2 - 7)}}{{2(5\sqrt 2 - 7)}} = 3}$$

№ 18 Есептеңіз: ${\dfrac{{25}}{{198}} \cdot \sqrt {\dfrac{{{{101}^2} - {{20}^2}}}{{125}}} - \dfrac{1}{{45}} \cdot \sqrt {{{115}^2} - {{110}^2}} }$

Шешуі: $${ = \dfrac{{25}}{{198}} \cdot \sqrt {\dfrac{{(101 - 20)(101 + 20)}}{{25 \cdot 5}}} - \dfrac{1}{{45}}\sqrt {(115 - 110)(115 + 110)} = }$$ $${ = \dfrac{{25}}{{198}}\sqrt {\dfrac{{81 \cdot 121}}{{25 \cdot 5}}} - \dfrac{1}{{45}}\sqrt {5 \cdot 225} = \dfrac{{25}}{{198}} \cdot \dfrac{{9 \cdot 11}}{{5\sqrt 5 }} - \dfrac{1}{{45}} \cdot 15 \cdot \sqrt 5 = }$$ $${ = \dfrac{5}{{2\sqrt 5 }} - \dfrac{{\sqrt 5 }}{3} = \dfrac{{\sqrt 5 \cdot \sqrt 5 }}{{2\sqrt 5 }} - \dfrac{{\sqrt 5 }}{3} = \sqrt 5 \left( {\dfrac{1}{2} - \dfrac{1}{3}} \right) = \dfrac{{\sqrt 5 }}{6}}$$

№ 19 Есептеңіз: $\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right) - \dfrac{1}{8} \cdot {3^{64}}$

Шешуі: $$ = \dfrac{{\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right)\left( {{3^4} + 1} \right)\left( {{3^8} + 1} \right)\left( {{3^{16}} + 1} \right)\left( {{3^{32}} + 1} \right)}}{8} - \dfrac{1}{8} \cdot {3^{64}}$$ $$\left( {{3^2} - 1} \right)\left( {{3^2} + 1} \right) = {3^4} - 1$$ $$\left( {{3^4} - 1} \right)\left( {{3^4} + 1} \right) = {3^8} - 1$$ $$\left( {{3^8} - 1} \right)\left( {{3^8} + 1} \right) = {3^{16}} - 1$$ $$\left( {{3^{16}} - 1} \right)\left( {{3^{16}} + 1} \right) = {3^{32}} - 1$$ $$\left( {{3^{32}} - 1} \right)\left( {{3^{32}} + 1} \right) = {3^{64}} - 1$$ $$\dfrac{{{3^{64}}}}{8} - \dfrac{1}{8} - \dfrac{{{3^{64}}}}{8} = - \dfrac{1}{8}$$

№ 20 Есептеңіз: $\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\left( {{2^{32}} + 1} \right)\left( {{2^{64}} + 1} \right) - \dfrac{1}{3} \cdot {2^{128}}$

Шешуі: $$ = \dfrac{{\left( {{2^2} - 1} \right)\left( {{2^2} + 1} \right)\left( {{2^4} + 1} \right)\left( {{2^8} + 1} \right)\left( {{2^{16}} + 1} \right)\left( {{2^{32}} + 1} \right)\left( {{2^{64}} + 1} \right)}}{3} - \dfrac{1}{3} \cdot {2^{128}}$$ $$\left( {{2^4} - 1} \right)\left( {{2^4} + 1} \right) = {2^8} - 1$$ $${\left( {{2^8} - 1} \right)\left( {{2^8} + 1} \right) = {2^{16}} - 1}$$ $${\left( {{2^{16}} - 1} \right)\left( {{2^{16}} + 1} \right) = {2^{32}} - 1}$$ $${\left( {{2^{32}} - 1} \right)\left( {{2^{32}} + 1} \right) = {2^{64}} - 1}$$ $$\left( {{2^{64}} - 1} \right)\left( {{2^{64}} + 1} \right) = {2^{128}} - 1$$ $$\dfrac{{{2^{128}}}}{3} - \dfrac{1}{3} - \dfrac{{{2^{128}}}}{3} = - \dfrac{1}{3}$$

 
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