Құрамында бүтін көрсеткішті дәрежелері бар сандық өрнектерді түрлендіріп, амалдарды орындаңыз.
№ 1 Есептеңіз: ${\left( {\frac{4}{3}} \right)^{ - 2}} \cdot {\left( {\frac{3}{4}} \right)^3}:{\left( { - \frac{2}{3}} \right)^{ - 3}} \cdot {\left( { - \frac{1}{2}} \right)^{ - 7}}$
Шешуі: $${\left( {\frac{4}{3}} \right)^{ - 2}} \cdot {\left( {\frac{3}{4}} \right)^3} = {\left( {\frac{3}{4}} \right)^2} \cdot {\left( {\frac{3}{4}} \right)^3} = {\left( {\frac{3}{4}} \right)^5}$$ $${\left( {\frac{3}{4}} \right)^5}:{\left( { - \left( {\frac{2}{3}} \right)} \right)^{ - 3}} \cdot {\left( { - \left( {\frac{1}{2}} \right)} \right)^{ - 7}} = {\left( {\frac{3}{4}} \right)^5}:{\left( {\frac{3}{2}} \right)^3} \cdot {2^7} = \frac{{{3^5}}}{{{2^{10}}}} \cdot \frac{{{2^3}}}{{{3^3}}} \cdot {2^7} = \frac{{{3^5}}}{{{3^3}}} = {3^2} = 9$$
№ 2 Есептеңіз: ${\left( {{{\left( {\frac{{{2^3}}}{5}} \right)}^{ - 1}} - 3 \cdot {{( - 2)}^{ - 3}} + {{\left( {\frac{7}{8}} \right)}^0}} \right)^{ - 1}}$
Шешуі: $${\left( {{{\left( {\frac{8}{5}} \right)}^{ - 1}} - 3 \cdot \frac{1}{{{{( - 2)}^3}}} + 1} \right)^{ - 1}} = {\left( {\frac{5}{8} + \frac{3}{8} + 1} \right)^{ - 1}} = {2^{ - 1}} = \frac{1}{2} = 0,5$$
№ 3 Есептеңіз: $\dfrac{{{{\left( {\frac{3}{2}} \right)}^{ - 3}} \cdot {{(3,375)}^{ - 1}}}}{{{{(2,25)}^{ - 2}} \cdot {{\left( {\frac{2}{3}} \right)}^{ - 1}}}}$
Шешуі: $${{{(3,375)}^{ - 1}} = {{\left( {3\frac{{375}}{{1000}}} \right)}^{ - 1}} = {{\left( {3\frac{3}{8}} \right)}^{ - 1}} = {{\left( {\frac{{27}}{8}} \right)}^{ - 1}} = {{\left( {{{\left( {\frac{3}{2}} \right)}^3}} \right)}^{ - 1}} = {{\left( {\frac{3}{2}} \right)}^{ - 3}}}$$ $${{{(2,25)}^{ - 2}} = {{\left( {2\frac{{25}}{{100}}} \right)}^{ - 2}} = {{\left( {2\frac{1}{4}} \right)}^{ - 2}} = {{\left( {\frac{9}{4}} \right)}^{ - 2}} = {{\left( {\frac{3}{2}} \right)}^{ - 4}}}$$ $${\frac{{{{\left( {\frac{3}{2}} \right)}^{ - 3}} \cdot {{\left( {\frac{3}{2}} \right)}^{ - 3}}}}{{{{\left( {\frac{3}{2}} \right)}^{ - 4}} \cdot \frac{3}{2}}} = {{\left( {\frac{3}{2}} \right)}^{ - 3}} = {{\left( {\frac{2}{3}} \right)}^3} = \frac{8}{{27}}}$$
№ 4 Есептеңіз: $\dfrac{{{{(0,4)}^{ - 2}} \cdot {{(2,5)}^{ - 4}}}}{{{{(0,16)}^{ - 5}} \cdot {{\left( {{{(6,25)}^{ - 3}}} \right)}^2}}}$
Шешуі: $$\frac{{{{(0,4)}^{ - 2}} \cdot {{(2,5)}^{ - 4}}}}{{{{(0,16)}^{ - 5}} \cdot {{\left( {{{(6,25)}^{ - 3}}} \right)}^2}}}$$ $${(0,4)^{ - 2}} = {\left( {\frac{4}{{10}}} \right)^{ - 2}} = {\left( {\frac{2}{5}} \right)^{ - 2}}$$ $${(2,5)^{ - 4}} = {\left( {2\frac{5}{{10}}} \right)^{ - 4}} = {\left( {2\frac{1}{2}} \right)^{ - 4}} = {\left( {\frac{5}{2}} \right)^{ - 4}} = {\left( {\frac{2}{5}} \right)^4}$$ $${(0,16)^{ - 5}} = {\left( {\frac{{16}}{{100}}} \right)^{ - 5}} = {\left( {\frac{4}{{25}}} \right)^{ - 5}} = {\left( {{{\left( {\frac{2}{5}} \right)}^2}} \right)^{ - 5}} = {\left( {\frac{2}{5}} \right)^{ - 10}}$$ $${\left( {{{(6,25)}^{ - 3}}} \right)^2} = {\left( {{{\left( {6\frac{{25}}{{100}}} \right)}^{ - 3}}} \right)^2} = {\left( {{{\left( {6\frac{1}{4}} \right)}^{ - 3}}} \right)^2} = {\left( {\frac{{25}}{4}} \right)^{ - 6}} = {\left( {{{\left( {\frac{5}{2}} \right)}^2}} \right)^{ - 6}} = {\left( {\frac{5}{2}} \right)^{ - 12}}$$ $$\frac{{{{\left( {\frac{2}{5}} \right)}^{ - 2}} \cdot {{\left( {\frac{2}{5}} \right)}^4}}}{{{{\left( {\frac{2}{5}} \right)}^{ - 10}} \cdot {{\left( {\frac{5}{2}} \right)}^{ - 12}}}} = \frac{{{{\left( {\frac{2}{5}} \right)}^2}}}{{{{\left( {\frac{2}{5}} \right)}^2}}} = 1$$
№ 5 Есептеңіз: ${( - 2,2)^5} \cdot {\left( {2\frac{3}{{11}}} \right)^4}$
Шешуі: $${{{( - 2,2)}^5} = {{\left( { - 2\frac{2}{{10}}} \right)}^5} = {{\left( { - 2\frac{1}{5}} \right)}^5} = {{\left( { - \frac{{11}}{5}} \right)}^5}}$$ $${{{\left( {2\frac{3}{{11}}} \right)}^4} = {{\left( {\frac{{25}}{{11}}} \right)}^4}}$$ $${{{\left( { - \frac{{11}}{5}} \right)}^5} \cdot {{\left( {\frac{{25}}{{11}}} \right)}^4} = - \frac{{11}}{{{5^5}}} \cdot \frac{{{5^8}}}{{{{11}^4}}} = - 11 \cdot {5^3} = - 1375}$$
№ 6 Есептеңіз: ${\left( {20 \cdot {2^4} - 12 \cdot {2^3} - 48 \cdot {2^2}} \right)^2}:{( - 8)^3}$
Шешуі: $${\left( {10 \cdot 2 \cdot {2^4} - 3 \cdot {2^2} \cdot {2^3} - 6 \cdot {2^3} \cdot {2^2}} \right)^2}:{( - 8)^3} = {\left( {10 \cdot {2^5} - 3 \cdot {2^5} - 6 \cdot {2^5}} \right)^2}:{( - 8)^3} = $$ $$ = {\left( {{2^5}(10 - 3 - 6)} \right)^2}:{( - 8)^3} = {\left( {{2^5}} \right)^2}:{\left( { - \left( {{2^3}} \right)} \right)^3} = - {2^{10}}:{2^9} = - 2$$
№ 7 Есептеңіз: $\left( {75 \cdot {5^2} + 35 \cdot {5^3}} \right):(20 \cdot 25 \cdot 125 - 625 \cdot 75)$
Шешуі: $$\left( {3 \cdot {5^2} \cdot {5^2} + 7 \cdot 5 \cdot {5^3}} \right):(20 \cdot 5 \cdot 625 - 625 \cdot 75) = $$ $$\left( {3 \cdot {5^4} + 7 \cdot {5^4}} \right):(625(100 - 75)) = {5^4}(3 + 7):625 \cdot 25 = \frac{{10 \cdot {5^4}}}{{{5^4} \cdot 25}} = \frac{{10}}{{25}} = \frac{2}{5}$$
№ 8 Есептеңіз: $\dfrac{{{{\left( {\frac{3}{{{2^3}}}} \right)}^{ - 1}} \cdot {{(1,5)}^3} + {{\left( { - \frac{1}{3}} \right)}^{ - 1}}}}{{{{(0,2)}^{ - 1}} + {{( - 2,3)}^0}}}$
Шешуі: $$\frac{{\frac{{{2^3}}}{3} \cdot {{\left( {\frac{3}{2}} \right)}^3} - 3}}{{{{\left( {\frac{1}{5}} \right)}^{ - 1}} + 1}} = \frac{{\frac{{{2^3}}}{3} \cdot \frac{{{3^3}}}{{{2^3}}} - 3}}{{5 + 1}} = \frac{{9 - 3}}{6} = \frac{6}{6} = 1$$
№ 9 Есептеңіз: $\dfrac{{{{( - 1,2)}^0} + {{(0,5)}^{ - 1}}}}{{{{\left( {\frac{{{2^3}}}{5}} \right)}^{ - 1}} - 3 \cdot {{( - 2)}^{ - 3}} - {{(0,2)}^{ - 2}}}}$
Шешуі: $$\frac{{1 + {{\left( {\frac{1}{2}} \right)}^{ - 1}}}}{{{{\left( {\frac{8}{5}} \right)}^{ - 1}} - 3{{( - (2))}^{ - 3}} - {{\left( {\frac{1}{5}} \right)}^{ - 2}}}} = \frac{{1 + 2}}{{\frac{5}{8} + 3 \cdot {2^{ - 3}} - {5^2}}} = \frac{3}{{\frac{5}{8} + \frac{3}{8} - 25}} = \frac{3}{{ - 24}} = - \frac{1}{8} = - 0,125$$
№ 10 Есептеңіз: ${\left( {6 - 4 \cdot {{\left( {\frac{5}{{16}}} \right)}^0}} \right)^{ - 2}} + {\left( {\frac{2}{3}} \right)^{ - 1}} - \frac{3}{{2\sqrt 2 }} \cdot \cos \frac{{3\pi }}{4}$
Шешуі: $${(6 - 4 \cdot 1)^{ - 2}} + \frac{3}{2} - \frac{3}{{2\sqrt 2 }} \cdot \left( { - \frac{{\sqrt 2 }}{2}} \right) = {2^{ - 2}} + \frac{3}{2} + \frac{3}{4} = \frac{1}{{{2^2}}} + \frac{3}{2} + \frac{3}{4} = $$ $$ = \frac{{10}}{4} = \frac{5}{2} = 2,5$$
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Есеп шешімдерінің авторы Күдебаева Ғалия Алмасқызы