Бүтін көрсеткішті дәрежесі бар сандық өрнектерді түрлендіру

Құрамында бүтін көрсеткіші бар сандық өрнектерді түрлендіру

1.13. Есептеңіз:

1) $\dfrac{2^{-2} \cdot 5^3 \cdot 10^{-4}}{2^{-3} \cdot 5^2 \cdot 10^{-5}}$

Шешуі

$$\frac{{{2^{ - 2}} \cdot {5^3} \cdot {{10}^{ - 4}}}}{{{2^{ - 3}} \cdot {5^2} \cdot {{10}^{ - 5}}}} = {2^{ - 2 + 3}} \cdot {5^{3 - 2}} \cdot {10^{ - 4 + 5}} = 2 \cdot 5 \cdot 10 = 100$$

2) $\dfrac{\left(2 \cdot(-3)^{-2}\right)^{-1} \cdot(-3)^{-2}}{4^{-1} \cdot(-3)^{-1} \cdot\left((-3)^{-2}\right)^{-1}}$

Шешуі

$$\frac{{{{\left( {2 \cdot {{( - 3)}^{ - 2}}} \right)}^{ - 1}} \cdot {{( - 3)}^{ - 2}}}}{{{4^{ - 1}} \cdot {{( - 3)}^{ - 1}} \cdot {{\left( {{{( - 3)}^{ - 2}}} \right)}^{ - 1}}}} = \frac{{{2^{ - 1}} \cdot {{( - 3)}^2} \cdot {{( - 3)}^{ - 2}}}}{{{2^{ - 2}} \cdot {{( - 3)}^{ - 1}} \cdot {{( - 3)}^2}}} = \frac{{{2^{ - 1}}}}{{{2^{ - 2}} \cdot ( - 3)}} = 2 \cdot \left( { - \frac{1}{3}} \right) = - \frac{2}{3}$$

3) $\dfrac{(1,5)^{-3} \cdot(3,375)^{-1}}{(2,25)^{-2} \cdot\left(\frac{2}{3}\right)^{-1}}$

Шешуі

$$\frac{{{{(1,5)}^{ - 3}} \cdot {{(3,375)}^{ - 1}}}}{{{{(2,25)}^{ - 2}} \cdot {{\left( {\frac{2}{3}} \right)}^{ - 1}}}} = \frac{{{{\left( {\frac{3}{2}} \right)}^{ - 3}} \cdot {{\left( {3\frac{3}{8}} \right)}^{ - 1}}}}{{{{\left( {2\frac{1}{4}} \right)}^{ - 2}} \cdot {{\left( {\frac{2}{3}} \right)}^{ - 1}}}} = \frac{{{{\left( {\frac{3}{2}} \right)}^{ - 3}} \cdot {{\left( {\frac{{27}}{8}} \right)}^{ - 1}}}}{{{{\left( {\frac{9}{4}} \right)}^{ - 2}} \cdot {{\left( {\frac{2}{3}} \right)}^{ - 1}}}} = $$ $$ = \frac{{{{\left( {\frac{3}{2}} \right)}^{ - 3}} \cdot {{\left( {\frac{3}{2}} \right)}^{ - 3}}}}{{{{\left( {\frac{3}{2}} \right)}^{ - 4}} \cdot \frac{3}{2}}} = {\left( {\frac{3}{2}} \right)^{ - 3}} = {\left( {\frac{2}{3}} \right)^3} = \frac{8}{{27}}$$

4) $\dfrac{(0,26)^0-(0,2)^{-1}}{\left(8: 5^3\right)^{-1} \cdot(0,4)^3+\left(-\frac{1}{2}\right)^{-1}}$

Шешуі

$$\frac{{{{(0,26)}^0} - {{(0,2)}^{ - 1}}}}{{{{\left( {8:{5^3}} \right)}^{ - 1}} \cdot {{(0,4)}^3} + {{\left( { - \frac{1}{2}} \right)}^{ - 1}}}} = \frac{{1 - 5}}{{\frac{{125}}{8} \cdot {{\left( {\frac{2}{5}} \right)}^3} + ( - 2)}} = \frac{{ - 4}}{{\frac{{125}}{8} \cdot \frac{8}{{125}} - 2}} = \frac{{ - 4}}{{1 - 2}} = 4$$

5) $\dfrac{40^4}{5^2 \cdot 2^{11}}+0,2^6 \cdot 5^6$

Шешуі

$$\frac{{{{40}^4}}}{{{5^2} \cdot {2^{11}}}} + {0,2^6} \cdot {5^6} = \frac{{{{\left( {{2^3} \cdot 5} \right)}^4}}}{{{5^2} \cdot {2^{11}}}} + {(0,2 \cdot 5)^6} = \frac{{{2^{12}} \cdot {5^4}}}{{{5^2} \cdot {2^{11}}}} + {1^6} = 2 \cdot {5^2} + 1 = 51$$

6) $12^6 \cdot 27^{-3} \cdot\left(\dfrac{1}{18}\right)^{-2} \cdot 4^{-5}$

Шешуі

$${12^6} \cdot {27^{ - 3}} \cdot {\left( {\frac{1}{{18}}} \right)^{ - 2}} \cdot {4^{ - 5}} = {\left( {{2^2} \cdot 3} \right)^6} \cdot {\left( {{3^3}} \right)^{ - 3}} \cdot {\left( {2 \cdot {3^2}} \right)^2} \cdot {\left( {{2^2}} \right)^{ - 5}} = $$ $$ = {2^{12}} \cdot {3^6} \cdot {3^{ - 9}} \cdot {2^2} \cdot {3^4} \cdot {2^{ - 10}} = {2^4} \cdot 3 = 48$$

7) $\left(\dfrac{1}{3}\right)^{-10} \cdot 27^{-3}+0,2^{-4} \cdot 25^{-2}+\left(64^{-\frac{1}{9}}\right)^{-3}$

Шешуі

$${\left( {\frac{1}{3}} \right)^{ - 10}} \cdot {27^{ - 3}} + {0,2^{ - 4}} \cdot {25^{ - 2}} + {\left( {{{64}^{ - \frac{1}{9}}}} \right)^{ - 3}} = {3^{10}} \cdot {3^{ - 9}} + {5^4} \cdot {5^{ - 4}} + {\left( {{2^6}} \right)^{\frac{1}{3}}} = 3 + {5^0} + {2^2} = 8$$

8) $0,3^{-3}+\left(\dfrac{3}{7}\right)^{-1}+(-0,5)^{-2} \cdot \dfrac{3}{4}+(-1)^8 \cdot 6$

Шешуі

$${0,3^{ - 3}} + {\left( {\frac{3}{7}} \right)^{ - 1}} + {( - 0,5)^{ - 2}} \cdot \frac{3}{4} + {( - 1)^8} \cdot 6 = \frac{{1000}}{{27}} + \frac{7}{3} + 4 \cdot \frac{3}{4} + 6 = 37\frac{1}{{27}} + 2\frac{1}{3} + 9 = 48\frac{{10}}{{27}}$$

9) $\dfrac{128 \cdot 8^{-1} \cdot 3^6+6^8}{4^2 \cdot 9^3 \cdot\left(\frac{1}{5}\right)^{-1}}:\left(\dfrac{1}{29}\right)^{-1}$

Шешуі

$$\frac{{128 \cdot {8^{ - 1}} \cdot {3^6} + {6^8}}}{{{4^2} \cdot {9^3} \cdot {{\left( {\frac{1}{5}} \right)}^{ - 1}}}}:{\left( {\frac{1}{{29}}} \right)^{ - 1}} = \frac{{{2^7} \cdot {2^{ - 3}} \cdot {3^6} + {{(2 \cdot 3)}^8}}}{{{2^4} \cdot {3^6} \cdot 5}}:29 = \frac{{{2^4} \cdot {3^6} + {2^8} \cdot {3^8}}}{{{2^4} \cdot {3^6} \cdot 5}} \cdot \frac{1}{{29}} = $$ $$ = \frac{{{2^4} \cdot {3^6}\left( {1 + {2^4} \cdot {3^2}} \right)}}{{{2^4} \cdot {3^6} \cdot 5}} \cdot \frac{1}{{29}} = \frac{{1 + 16 \cdot 9}}{{5 \cdot 29}} = \frac{{145}}{{145}} = 1$$

10) $\dfrac{6 \cdot\left(\frac{1}{15}\right)^{-1}}{\left(2^{-3}\right)^2 \cdot\left(\frac{1}{8}\right)^{-4} \cdot \sqrt{16^{-1}}+243 \cdot\left(-\frac{1}{3}\right)^5}$

Шешуі

$$\frac{{6 \cdot {{\left( {\frac{1}{{15}}} \right)}^{ - 1}}}}{{{{\left( {{2^{ - 3}}} \right)}^2} \cdot {{\left( {\frac{1}{8}} \right)}^{ - 4}} \cdot \sqrt {{{16}^{ - 1}}} + 243 \cdot {{\left( { - \frac{1}{3}} \right)}^5}}} = \frac{{6 \cdot 15}}{{{2^{ - 6}} \cdot {2^{12}} \cdot {2^{ - 2}} + {3^5} \cdot {{( - 3)}^{ - 5}}}} = $$ $$ = \frac{{90}}{{{2^4} - {3^0}}} = \frac{{90}}{{16 - 1}} = 6$$



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Пікір қалдыру

Сіздің электронды почтаңыз жарияланбайды, Міндетті жолдарды толтырып шығыңыз.