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Câu hỏi: Tính:
\(a) {25^3}:{5^2};\)
\(b) \displaystyle {\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6};\)
\(c) \displaystyle 3 - {\left( { - {6 \over 7}} \right)^0} + {\left( {{1 \over 2}} \right)^2}:2\)
\(a) {25^3}:{5^2};\)
\(b) \displaystyle {\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6};\)
\(c) \displaystyle 3 - {\left( { - {6 \over 7}} \right)^0} + {\left( {{1 \over 2}} \right)^2}:2\)
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
\(a) {25^3}:{5^2} = {25^3}:25 = {25^2} = 625\)
\(\displaystyle b) {\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6}\)
\(\displaystyle = {\left( {{3 \over 7}} \right)^{21}}:{\left[ {{{\left( {{3 \over 7}} \right)}^2}} \right]^6} \)
\(\displaystyle = {\left( {{3 \over 7}} \right)^{21}}:{\left( {{3 \over 7}} \right)^{12}} \)
\(\displaystyle = {\left( {{3 \over 7}} \right)^9} = {{19683} \over {40353607}}\)
\(c) \displaystyle3 - {\left( { - {6 \over 7}} \right)^0} + {\left( {{1 \over 2}} \right)^2}:2\)
\(\displaystyle= 3 - 1 + {\left( {{1 \over 2}} \right)^2}.\frac{1}{2}\)
\(\displaystyle = 2 + {1 \over 8} = 2{1 \over 8}\)
\({x^n} = \underbrace {x \ldots x}_{n thừa số}\) 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
\(a) {25^3}:{5^2} = {25^3}:25 = {25^2} = 625\)
\(\displaystyle b) {\left( {{3 \over 7}} \right)^{21}}:{\left( {{9 \over {49}}} \right)^6}\)
\(\displaystyle = {\left( {{3 \over 7}} \right)^{21}}:{\left[ {{{\left( {{3 \over 7}} \right)}^2}} \right]^6} \)
\(\displaystyle = {\left( {{3 \over 7}} \right)^{21}}:{\left( {{3 \over 7}} \right)^{12}} \)
\(\displaystyle = {\left( {{3 \over 7}} \right)^9} = {{19683} \over {40353607}}\)
\(c) \displaystyle3 - {\left( { - {6 \over 7}} \right)^0} + {\left( {{1 \over 2}} \right)^2}:2\)
\(\displaystyle= 3 - 1 + {\left( {{1 \over 2}} \right)^2}.\frac{1}{2}\)
\(\displaystyle = 2 + {1 \over 8} = 2{1 \over 8}\)