Giải Bài 24 trang 18 sách bài tập toán 7 tập 1 - Cánh diều

Đề bài

So sánh:

a) \({\left( { - {\rm{ }}0,1} \right)^2}.{\left( { - {\rm{ }}0,1} \right)^4}\) và \({\left[ {{{\left( { - {\rm{ }}0,1} \right)}^3}} \right]^2}\);      

b) \({\left( {\dfrac{1}{2}} \right)^8}:{\left( {\dfrac{1}{2}} \right)^2}\) và \({\left( {\dfrac{1}{2}} \right)^3}.{\left( {\dfrac{1}{2}} \right)^3}\);

c) \({9^8}:{27^3}\) và \({3^2}{.3^5}\);                                         

d) \({\left( {\dfrac{1}{4}} \right)^7}.0,25\) và \({\left[ {{{\left( {\dfrac{1}{4}} \right)}^2}} \right]^4}\);

e) \({\left[ {{{\left( { - {\rm{ }}0,7} \right)}^2}} \right]^3}\) và \({\left[ {{{\left( {0,7} \right)}^3}} \right]^2}\).

Lời giải chi tiết

a) Ta có:

\({\left( { - {\rm{ }}0,1} \right)^2}.{\left( { - {\rm{ }}0,1} \right)^4} = {\left( { - {\rm{ }}0,1} \right)^{2 + 4}} = {( - {\rm{ }}0,1)^6}\) ; \({\left[ {{{\left( { - {\rm{ }}0,1} \right)}^3}} \right]^2} = {\left( { - {\rm{ }}0,1} \right)^{3.2}} = {\left( { - {\rm{ }}0,1} \right)^6}\)

Vậy \({\left( { - {\rm{ }}0,1} \right)^2}.{\left( { - {\rm{ }}0,1} \right)^4}\) = \({\left[ {{{\left( { - {\rm{ }}0,1} \right)}^3}} \right]^2}\).

b) Ta có:

     \({\left( {\dfrac{1}{2}} \right)^8}:{\left( {\dfrac{1}{2}} \right)^2} = {\left( {\dfrac{1}{2}} \right)^{8 - 2}} = {\left( {\dfrac{1}{2}} \right)^6}\) ; \({\left( {\dfrac{1}{2}} \right)^3}.{\left( {\dfrac{1}{2}} \right)^3} = {\left( {\dfrac{1}{2}} \right)^{3 + 3}} = {\left( {\dfrac{1}{2}} \right)^6}\)

Vậy \({\left( {\dfrac{1}{2}} \right)^8}:{\left( {\dfrac{1}{2}} \right)^2}\) = \({\left( {\dfrac{1}{2}} \right)^3}.{\left( {\dfrac{1}{2}} \right)^3}\).

c) Ta có:

     \({9^8}:{27^3} = {\left( {{3^2}} \right)^8}:{\left( {{3^3}} \right)^3} =3^{2.8}:3^{3.3}= {3^{16}}:{3^9} = {3^{16 - 9}} = {3^7};\\ {3^2}{.3^5} = {3^{2 + 5}} = {3^7}\)

Vậy \({9^8}:{27^3}={3^2}{.3^5}\).                

d) Ta có:

     \({\left( {\dfrac{1}{4}} \right)^7}.0,25 = {\left( {\dfrac{1}{4}} \right)^7}.\left( {\dfrac{1}{4}} \right) = {\left( {\dfrac{1}{4}} \right)^{7 + 1}} = {\left( {\dfrac{1}{4}} \right)^8}\) ; \({\left[ {{{\left( {\dfrac{1}{4}} \right)}^2}} \right]^4} = {\left( {\dfrac{1}{4}} \right)^{2.4}} = {\left( {\dfrac{1}{4}} \right)^8}\)

Vậy \({\left( {\dfrac{1}{4}} \right)^7}.0,25\) = \({\left[ {{{\left( {\dfrac{1}{4}} \right)}^2}} \right]^4}\).

e) Ta có:

     \({\left[ {{{\left( { - 0,7} \right)}^2}} \right]^3} = {\left[ {{{\left( {0,7} \right)}^2}} \right]^3} = {(0,7)^{2.3}} = {(0,7)^6}\) ; \({\left[ {{{\left( {0,7} \right)}^3}} \right]^2} = {(0,7)^{3.2}} = {(0,7)^6}\).

Vậy \({\left[ {{{\left( { - 0,7} \right)}^2}} \right]^3}\) = \({\left[ {{{\left( {0,7} \right)}^3}} \right]^2}\).