Google
Câu hỏi: Viết các biểu thức số sau dưới dạng \({{\rm{a}}^n}(a \in\mathbb Q,n \in\mathbb N)\):
a) \(\displaystyle {9.3^3}.{1 \over {81}}{.3^2}\)
b) \(\displaystyle {4.2^5}:\left( {{2^3}.{1 \over {16}}} \right)\)
c) \(\displaystyle {3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2}\)
d) \(\displaystyle {\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2}\)
a) \(\displaystyle {9.3^3}.{1 \over {81}}{.3^2}\)
b) \(\displaystyle {4.2^5}:\left( {{2^3}.{1 \over {16}}} \right)\)
c) \(\displaystyle {3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2}\)
d) \(\displaystyle {\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2}\)
Phương pháp giải
\({x^n} = \underbrace {x \ldots x}_{n thừa số}\) với \(( x ∈\mathbb Q, n ∈\mathbb N, n> 1)\)
Nếu \(x = \dfrac{a}{b}\) thì \({x^n} = {\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{a^n}}}{{{b^n}}}\)
\({x^m}.{x^n} = {x^{m + n}}\) (\( x ∈\mathbb Q, m,n ∈\mathbb N\))
\({x^m}:{x^n} = {x^{m - n}}\) (\(x ≠ 0, m ≥ n\))
\({\left( {{x^m}} \right)^n} = {x^{m.n}}\)
Quy ước:
\(\eqalign{
& {a^o} = 1 \left( {a \in {\mathbb N^*}} \right) \cr
& {x^o} = 1 \left( {x \in\mathbb Q, x \ne 0} \right) \cr} \)
Lời giải chi tiết
a) \(\displaystyle {9.3^3}.{1 \over {81}}{.3^2} = \left( {{3^2}{{.3}^3}{{.3}^2}} \right).{1 \over {{3^4}}}\)\(\displaystyle = {{{3^{2+3+2}}} \over {{3^4}}} \) \(\displaystyle = {{{3^7}} \over {{3^4}}} = {3^3}\)
b) \(\displaystyle {4.2^5}:\left( {{2^3}.{1 \over {16}}} \right) = {2^2}{.2^5}:\left( {{2^3}.{1 \over {{2^4}}}} \right) \)\( \displaystyle = {2^7}:{1 \over 2} = {2^7}.2 = {2^8}\)
c) \(\displaystyle {3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2} = {3^2}{.2^5}.{{{2^2}} \over {{3^2}}} \)\(\displaystyle = {2^5}.{{{2^2}} \over {{3^2}}}.{3^2} \)
\(\displaystyle = {{2^5}{{.2}^2}} = {2^{5+2}} = {2^7}\)
d) \(\displaystyle {\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2} = \left( {{1 \over {{3^2}}}.{1 \over 3}} \right).{\left( {{3^2}} \right)^2}\)\( \displaystyle = {1 \over {{3^3}}}{.3^4} = 3^1=3\).
\({x^n} = \underbrace {x \ldots x}_{n thừa số}\) với \(( x ∈\mathbb Q, n ∈\mathbb N, n> 1)\)
Nếu \(x = \dfrac{a}{b}\) thì \({x^n} = {\left( {\dfrac{a}{b}} \right)^n} = \dfrac{{{a^n}}}{{{b^n}}}\)
\({x^m}.{x^n} = {x^{m + n}}\) (\( x ∈\mathbb Q, m,n ∈\mathbb N\))
\({x^m}:{x^n} = {x^{m - n}}\) (\(x ≠ 0, m ≥ n\))
\({\left( {{x^m}} \right)^n} = {x^{m.n}}\)
Quy ước:
\(\eqalign{
& {a^o} = 1 \left( {a \in {\mathbb N^*}} \right) \cr
& {x^o} = 1 \left( {x \in\mathbb Q, x \ne 0} \right) \cr} \)
Lời giải chi tiết
a) \(\displaystyle {9.3^3}.{1 \over {81}}{.3^2} = \left( {{3^2}{{.3}^3}{{.3}^2}} \right).{1 \over {{3^4}}}\)\(\displaystyle = {{{3^{2+3+2}}} \over {{3^4}}} \) \(\displaystyle = {{{3^7}} \over {{3^4}}} = {3^3}\)
b) \(\displaystyle {4.2^5}:\left( {{2^3}.{1 \over {16}}} \right) = {2^2}{.2^5}:\left( {{2^3}.{1 \over {{2^4}}}} \right) \)\( \displaystyle = {2^7}:{1 \over 2} = {2^7}.2 = {2^8}\)
c) \(\displaystyle {3^2}{.2^5}.{\left( {{2 \over 3}} \right)^2} = {3^2}{.2^5}.{{{2^2}} \over {{3^2}}} \)\(\displaystyle = {2^5}.{{{2^2}} \over {{3^2}}}.{3^2} \)
\(\displaystyle = {{2^5}{{.2}^2}} = {2^{5+2}} = {2^7}\)
d) \(\displaystyle {\left( {{1 \over 3}} \right)^2}.{1 \over 3}{.9^2} = \left( {{1 \over {{3^2}}}.{1 \over 3}} \right).{\left( {{3^2}} \right)^2}\)\( \displaystyle = {1 \over {{3^3}}}{.3^4} = 3^1=3\).