# Toán Lớp 7 Bài 2

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:

$$\begin{array}{l}{x^m}.{x^n} = {x^{m + n}}\left( {m,n \in \mathbb{N}} \right)\\{x^m}:{x^n} = {x^{m - n}}\left( {x \ne 0;m \ge n;\,m,n \in \mathbb{N}} \right)\end{array}$$

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 $${\left( { - \frac{3}{2}} \right)^5}:{\left( { - \frac{3}{2}} \right)^3}$$ = $${\left( {\frac{3}{2}} \right)^2}$$.