Giải bài 2 trang 20 SGK Toán 7 tập 1 – Cánh diều

Đề bài

So sánh:

a) \({( – 2)^4} \cdot {( – 2)^5}\) và \({( – 2)^{12}}:{( – 2)^3}\);

b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) và \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)

c) \({(0,3)^8}:{(0,3)^2}\) và \({\left[ {{{(0,3)}^2}} \right]^3}\);

d) \({\left( { – \frac{3}{2}} \right)^5}:{\left( { – \frac{3}{2}} \right)^3}\) và \({\left( {\frac{3}{2}} \right)^2}\).

Lời giải chi tiết

a) \({( – 2)^4} \cdot {( – 2)^5} = {\left( { – 2} \right)^{4 + 5}} = {\left( { – 2} \right)^9}\)

 \({( – 2)^{12}}:{( – 2)^3} = {\left( { – 2} \right)^{12 – 3}} = {\left( { – 2} \right)^9}\)

Vậy \({( – 2)^4} \cdot {( – 2)^5}\) = \({( – 2)^{12}}:{( – 2)^3}\);

b) \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6} = {\left( {\frac{1}{2}} \right)^{2 + 6}} = {\left( {\frac{1}{2}} \right)^8}\)

\({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2} = {\left( {\frac{1}{2}} \right)^{4.2}} = {\left( {\frac{1}{2}} \right)^8}\)

Vậy \({\left( {\frac{1}{2}} \right)^2} \cdot {\left( {\frac{1}{2}} \right)^6}\) = \({\left[ {{{\left( {\frac{1}{2}} \right)}^4}} \right]^2}\)

c) \({(0,3)^8}:{(0,3)^2} = {\left( {0,3} \right)^{8 – 2}} = {\left( {0,3} \right)^6}\)

\({\left[ {{{(0,3)}^2}} \right]^3} = {\left( {0,3} \right)^{2.3}} = {\left( {0,3} \right)^6}\)

Vậy \({(0,3)^8}:{(0,3)^2}\)= \({\left[ {{{(0,3)}^2}} \right]^3}\).

d) \({\left( { – \frac{3}{2}} \right)^5}:{\left( { – \frac{3}{2}} \right)^3} = {\left( { – \frac{3}{2}} \right)^{5 – 3}} = {\left( { – \frac{3}{2}} \right)^2} = {\left( {\frac{3}{2}} \right)^2}\)

Vậy \({\left( { – \frac{3}{2}} \right)^5}:{\left( { – \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).