Bài tập Lũy thừa của một số hữu tỉ có đáp án

$$\begin{gathered} {\left(\frac{-2}{3}\right)}^3=\frac{{\left(-2\right)}^3}{3^3}=\frac{-8}{27} \\ {\left(\frac{-3}{5}\right)}^2=\frac{{\left(-3\right)}^2}{5^2}=\frac{9}{25} \\ {\left(-0,5\right)}^3={\left(\frac{-1}{2}\right)}^3=\frac{{\left(-1\right)}^3}{2^3}=\frac{-1}{8} \\ {\left(-0,5\right)}^2={\left(\frac{-1}{2}\right)}^2=\frac{{\left(-1\right)}^2}{2^2}=\frac{1}{4} \\ {\left(37,57\right)}^0=1 \\ {\left(3,57\right)}^1=3,57 \end{gathered}$$

a) Ta có:

${\left(\frac{1}{3}\right)}^2.{\left(\frac{1}{3}\right)}^2=\frac{1^2}{3^2}.\frac{1^2}{3^2}=\frac{1^4}{3^2{.3}^2}=\frac{1^4}{3^4}=\frac{1}{81}={\left(\frac{1}{3}\right)}^4$

Vậy “?” điền 4.

b) Ta có:

a) Ta có: ${\left[{\left(-2\right)}^2\right]}^3=$4³ = 64 và  = 64

Do đó, ${\left[{\left(-2\right)}^2\right]}^3={\left(-2\right)}^6$

b) $Ta có {\left[{\left(\frac{1}{2}\right)}^2\right]}^2={\left(\frac{1^2}{2^2}\right)}^2={\left(\frac{1}{4}\right)}^2=\frac{1^2}{4^2}=\frac{1}{16}$

${\left(\frac{1}{2}\right)}^4=\frac{1^4}{2^4}=\frac{1}{16}$

a) ${\left[{\left(\frac{-2}{3}\right)}^2\right]}^5={\left(\frac{-2}{3}\right)}^{2.5}={\left(\frac{-2}{3}\right)}^{10}$ 

Vậy "?" là 10.

b) ${\left[{\left(0,4\right)}^3\right]}^3={\left(0,4\right)}^{3.3}={\left(0,4\right)}^9$

Vậy "?" là 9

c) ${\left[{\left(7,31\right)}^3\right]}^0={\left(7,31\right)}^{3.0}={\left(7,31\right)}^0=1$

Vậy "?" là 1

$$\begin{gathered} a) {\left(\frac{-1}{2}\right)}^5=\frac{{\left(-1\right)}^5}{2^5}=\frac{-1}{32}; \\ {\left(\frac{-2}{3}\right)}^4=\frac{{\left(-2\right)}^4}{3^4}=\frac{16}{81}; \\ {\left(-2\frac{1}{4}\right)}^3={\left(-\frac{2.4+1}{4}\right)}^3={\left(\frac{-9}{4}\right)}^3=\frac{{\left(-9\right)}^3}{4^3}=\frac{-729}{64}; \\ {\left(-\mathrm{0,3}\right)}^5={\left(\frac{-3}{10}\right)}^5=\frac{{\left(-3\right)}^5}{{10}^5}=\frac{-243}{100000}; \\ {\left(-25,7\right)}^0=1; \end{gathered}$$

$$\begin{gathered} b) {\left(-\frac{1}{3}\right)}^2=\frac{{\left(-1\right)}^2}{3^2}=\frac{1}{9}; \\ {\left(-\frac{1}{3}\right)}^3=\frac{{\left(-1\right)}^3}{3^3}=\frac{-1}{27}; \\ {\left(-\frac{1}{3}\right)}^4=\frac{{\left(-1\right)}^4}{3^4}=\frac{1}{81}; \\ {\left(-\frac{1}{3}\right)}^5=\frac{{\left(-1\right)}^5}{3^5}=\frac{-1}{243}. \end{gathered}$$

$$\begin{gathered} a) x:{\left(\frac{-1}{2}\right)}^3=\frac{-1}{2} \\ x:\frac{{\left(-1\right)}^3}{2^3}=\frac{-1}{2} \\ x:\frac{-1}{8}=\frac{-1}{2} \\ x=\frac{-1}{2}.\frac{-1}{8} \\ x=\frac{1}{16} \end{gathered}$$

Vậy $x=\frac{1}{16}$

$$\begin{gathered} b) x.{\left(\frac{3}{5}\right)}^7={\left(\frac{3}{5}\right)}^9 \\ x={\left(\frac{3}{5}\right)}^9:{\left(\frac{3}{5}\right)}^7 \\ x={\left(\frac{3}{5}\right)}^{9-7} \\ x={\left(\frac{3}{5}\right)}^2=\frac{9}{25} \\ Vậy x=\frac{9}{25} \end{gathered}$$

$$\begin{gathered} c) {\left(\frac{-2}{3}\right)}^{11}:x={\left(\frac{-2}{3}\right)}^9 \\ x={\left(\frac{-2}{3}\right)}^{11}:{\left(\frac{-2}{3}\right)}^9 \\ x={\left(\frac{-2}{3}\right)}^{11-9} \\ x={\left(\frac{-2}{3}\right)}^2=\frac{4}{9} \\ Vậy x=\frac{4}{9} \end{gathered}$$

$$\begin{gathered} d) x.{\left(0,25\right)}^6={\left(\frac{1}{4}\right)}^8 \\ x.{\left(\frac{1}{4}\right)}^6={\left(\frac{1}{4}\right)}^8 \\ x={\left(\frac{1}{4}\right)}^8:{\left(\frac{1}{4}\right)}^6 \\ x={\left(\frac{1}{4}\right)}^{8-6} \\ x={\left(\frac{1}{4}\right)}^2=\frac{1}{16} \\ Vậy x=\frac{1}{16} \end{gathered}$$

$$\begin{gathered} {\left(\mathrm{0,25}\right)}^8={\left(\frac{25}{100}\right)}^8={\left(\frac{5^2}{{10}^2}\right)}^8={\left[{\left(\frac{5}{10}\right)}^2\right]}^8 \\ ={\left[{\left(\mathrm{0,5}\right)}^2\right]}^8={\left(\mathrm{0,5}\right)}^{2.8}={\left(\mathrm{0,5}\right)}^{16}. \\ {\left(\mathrm{0,125}\right)}^4={\left(\frac{125}{1000}\right)}^4={\left(\frac{5^3}{{10}^3}\right)}^4={\left[{\left(\frac{5}{10}\right)}^3\right]}^4 \\ ={\left[{\left(\mathrm{0,5}\right)}^3\right]}^4={\left(\mathrm{0,5}\right)}^{3.4}={\mathrm{0,5}}^{12}. \\ {\left(\mathrm{0,0625}\right)}^2={\left(\frac{625}{10000}\right)}^2={\left(\frac{5^4}{{10}^4}\right)}^2={\left[{\left(\frac{5}{10}\right)}^4\right]}^2 \\ ={\left[{\left(\mathrm{0,5}\right)}^4\right]}^2={\left(\mathrm{0,5}\right)}^{4.2}={\left(\mathrm{0,5}\right)}^8. \end{gathered}$$

$$\begin{gathered} a) \left[{\left(\frac{3}{7}\right)}^4.{\left(\frac{3}{7}\right)}^5\right]:{\left(\frac{3}{7}\right)}^7=\left[{\left(\frac{3}{7}\right)}^{4+5}\right]:{\left(\frac{3}{7}\right)}^7 \\ ={\left(\frac{3}{7}\right)}^9:{\left(\frac{3}{7}\right)}^7={\left(\frac{3}{7}\right)}^{9-7}={\left(\frac{3}{7}\right)}^2 \end{gathered}$$

$$\begin{gathered} b) \left[{\left(\frac{7}{8}\right)}^5:{\left(\frac{7}{8}\right)}^4\right].\frac{7}{8}=\left[{\left(\frac{7}{8}\right)}^{5-4}\right].\frac{7}{8} \\ ={\left(\frac{7}{8}\right)}^1.{\left(\frac{7}{8}\right)}^1={\left(\frac{7}{8}\right)}^{1+1}={\left(\frac{7}{8}\right)}^2 \end{gathered}$$

$$\begin{gathered} c) \left[{\left(0,6\right)}^3.{\left(0,6\right)}^8\right]:\left[{\left(0,6\right)}^7.{\left(0,6\right)}^2\right] \\ =\left[{\left(0,6\right)}^{3+8}\right]:\left[{\left(0,6\right)}^{7+2}\right] \\ ={\left(0,6\right)}^{11}:{\left(0,6\right)}^9 \\ ={\left(0,6\right)}^{11-9}={\left(0,6\right)}^2 \end{gathered}$$

a) ${\left(\frac{2}{5}+\frac{1}{2}\right)}^2={\left(\frac{4}{10}+\frac{5}{10}\right)}^2={\left(\frac{9}{10}\right)}^2=\frac{9^2}{{10}^2}=\frac{81}{100}$

b)$$\begin{gathered} {\left(\mathrm{0,75}-1\frac{1}{2}\right)}^3={\left(\frac{75}{100}-\frac{3}{2}\right)}^3 \\ ={\left(\frac{3}{4}-\frac{3}{2}\right)}^3={\left(\frac{3}{4}-\frac{6}{4}\right)}^3 \\ ={\left(\frac{-3}{4}\right)}^3=\frac{{\left(-3\right)}^3}{4^3}=\frac{-27}{64} \end{gathered}$$

 $$\begin{gathered} {\left(\frac{3}{5}\right)}^{15}:{\left(\mathrm{0,36}\right)}^5={\left(\frac{3}{5}\right)}^{15}:{\left(\frac{36}{100}\right)}^5 \\ ={\left(\frac{3}{5}\right)}^{15}:{\left(\frac{9}{25}\right)}^5=\frac{3^{15}}{5^{15}}:{\left(\frac{3^2}{5^2}\right)}^5 \\ ={\left(\frac{3}{5}\right)}^{15}:{\left(\frac{3}{2}\right)}^{2.5}={\left(\frac{3}{5}\right)}^{15}:{\left(\frac{3}{5}\right)}^{10} \\ ={\left(\frac{3}{5}\right)}^{15-10}={\left(\frac{3}{5}\right)}^5=\frac{3^5}{5^5}=\frac{243}{3125}. \end{gathered}$$

d) ${\left(1-\frac{1}{3}\right)}^8:{\left(\frac{4}{9}\right)}^3={\left(\frac{2}{3}\right)}^8:{\left[{\left(\frac{2}{3}\right)}^2\right]}^3$

$={\left(\frac{2}{3}\right)}^8:{\left(\frac{2}{3}\right)}^{2.3}={\left(\frac{2}{3}\right)}^8:{\left(\frac{2}{3}\right)}^6$

$={\left(\frac{2}{3}\right)}^{8-6}={\left(\frac{2}{3}\right)}^2$=$\frac{2^2}{3^2}=\frac{4}{9}.$

a) $\frac{4^3{.9}^7}{{27}^5{.8}^2}=\frac{{\left(2^2\right)}^3.{\left(3^2\right)}^7}{{\left(3^3\right)}^5.{\left(2^3\right)}^2}=\frac{2^6{.3}^{14}}{3^{15}{.2}^6}=\frac{1}{3}$

b) $\frac{{\left(-2\right)}^3.{\left(-2\right)}^7}{{3.4}^6}=\frac{{\left(-2\right)}^{3+7}}{3.{\left(2^2\right)}^6}$

$=\frac{{\left(-2\right)}^{10}}{{3.2}^{12}}=\frac{2^{10}}{{3.2}^{12}}=\frac{1}{{3.2}^2}=\frac{1}{12}$

c) $\frac{{\left(0,2\right)}^5.{\left(0,09\right)}^3}{{\left(0,2\right)}^7.{\left(0,3\right)}^4}=\frac{{\left(0,2\right)}^5.{\left(0,3^2\right)}^3}{{\left(0,2\right)}^7.{\left(0,3\right)}^4}$

$=\frac{{\left(0,2\right)}^5.{\left(0,3^2\right)}^3}{{\left(0,2\right)}^7.{\left(0,3\right)}^4}=\frac{{\left(0,2\right)}^5.{\left(0,3\right)}^6}{{\left(0,2\right)}^7.{\left(0,3\right)}^4}$

$=\frac{0,3^2}{0,2^2}={\left(\frac{0,3}{0,2}\right)}^2={\left(\frac{3}{2}\right)}^2=\frac{9}{4}$

d) $\frac{2^3+2^4+2^5}{7^2}=\frac{8+16+32}{49}=\frac{56}{49}=\frac{8}{7}$