a)\({\left( { - 2} \right)^2}.{\left( { - 2} \right)^3} = {\left( { - 2} \right)^{2 + 3}} = {\left( { - 2} \right)^5}\);

b)\({\left( { - 0,25} \right)^7}:{\left( { - 0,25} \right)^5} = {\left( { - 0,25} \right)^{7 - 5}} = {\left( { - 0,25} \right)^2} = {\left( {0,25} \right)^2}\);

c)\({\left( {\frac{3}{4}} \right)^4}.{\left( {\frac{3}{4}} \right)^3} = {\left( {\frac{3}{4}} \right)^{4 + 3}} = {\left( {\frac{3}{4}} \right)^7}.\)

sao bạn tự đăng tự giải thế? hay là bạn giải cho ai à?

ukm bn

a, \(\frac{1^5}{5}\).5⁵= (\(\frac{1}{5}\).5)⁵=1⁵=1

a) \(\left(\frac{1}{5}\right)^5.5^5=\left(\frac{1}{5}.5\right)^5=1^5=1\)

b) \(\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)

c) \(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)^4.1024=\frac{1^4}{4^4}.1024=\frac{1}{16}.1024=\frac{1024}{16}=58\)

a) \(\left(\frac{1}{5}\right)^5.5^5=\left(\frac{1}{5}.5\right)^5=1^5=1\)

b) \(\left(0,125\right)^3.512=\left(0,125\right)^3.8^3=\left(0,125.8\right)^3=1^3=1\)

c) \(\left(0,25\right)^4.1024=\text{[}\left(0,25\right)^2\text{]}^2.32^2=\left(\frac{1}{6}\right)^2.32^2=\left(\frac{1}{6}.32\right)^2=\left(\frac{32}{6}\right)^2=2^2=4\)

a)1

b)1

c)4

A) (1/5)^5 . 5^5 = 1/5^5 . 5^5 = 5^5 / 5^5 = 1.

B)(0,125)^3 . 512 = (1/8)^3 . 512 = 1/8^3 . 512 = 1/512 . 512 = 1.

C) (0,25)^4 . 1024 = (1/4)^4 . 1024 = 1/4^4 . 1024 = 1/256 . 1024 = 4.

a) \(\left(\frac{1}{5}\right)^5.5^5=\frac{1^5}{5^5}.5^5=1^5=1\)

b) \(\left(0,25\right)^4.1024=\left(\frac{1}{4}\right)\text{ }^4.2^{10}=\frac{1^4}{4^4}.2^{10}=\frac{1}{\left(2^2\right)^4}.2^{10}=\frac{1}{2^8}.2^{10}=2^2=4\)

c) \(\left(0,125\right)^3.512=\left(\frac{1}{8}\right)^3.2^9=\frac{1^3}{8^3}.2^9=\frac{1}{\left(2^3\right)^3}.2^9=\frac{1}{2^9}.2^9=1\)

a)

\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)                         

b)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ =  - 2 - \frac{5}{7}.\frac{7}{{15}}\\ =  - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)

c)

\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)             

d)

\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)

a)

\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\=  - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)                          

b)

\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)                  

d)

\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ =  - \frac{3}{4}\end{array}\).

\(\left(\dfrac{1}{7}\right)^7\cdot7^7=\left(\dfrac{1}{7}\cdot7\right)^7=1^7=1\\ \left(0,125\right)^3\cdot512=\left(0,125\right)^3\cdot8^3=\left(0,125\cdot8\right)^3=1^3=1\\ \left(0,25\right)^4\cdot1024=\left(0,25\right)^4\cdot256\cdot4=\left(0,25\right)^4\cdot4^4\cdot4=\left(0,25\cdot4\right)^4\cdot4=1^4\cdot4=4\)

a) \(\left(\dfrac{1}{7}\right)^7.7^7=\left(\dfrac{1}{7}.7\right)^7=1^7=1\)

b) \(\left(0.125\right)^3.512=\left(0.125\right)^3.8^3=\left(0.125\cdot8\right)^3=1^3=1\)

c) \(\left(0.25\right)^4.1024=\left(0.25\right)^4.4^5=\left(0.25\right)^4.4^4.4=\left(0.25.4\right)^4.4=1^4.4=1.4=4\)