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 \({\left( {\frac{3}{2}} \right)^2}\).
Phương pháp giải - Xem chi tiết
Thực hiện phép tính rồi so sánh:
\(\begin{array}{l}{x^m}.{x^n} = {x^{m + n}}\left( {m,n \in \mathbb{N}} \right)\\{x^m}:{x^n} = {x^{m - n}}\left( {x \ne 0;m \ge n;\,m,n \in \mathbb{N}} \right)\end{array}\)
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 \({\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}\) = \({\left( {\frac{3}{2}} \right)^2}\).
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