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a) \(\dfrac{2^{14}.3^{12}}{6^{11}}\)
\(=\dfrac{2^2.2^{12}.3^{12}}{6^{11}}\)
\(=\dfrac{4.6^{12}}{6^{11}}\)
\(=4.6\)
\(=24\)
\(\begin{array}{l}a){( - 2)^3}.{( - 2)^4} = {( - 2)^{3 + 4}} = {( - 2)^7}\\b){(0,25)^7}:{(0,25)^3} = {(0,25)^{7 - 3}} = {(0,25)^4}\end{array}\)
`(1 1/4)^10 . (2/5)^20`
`=(5/4)^10 . (2/5)^20`
`=(5^10 .2^20)/(4^10 .5^20)`
`=(5^10 .4^10)/(4^10 .5^20)`
`=1/(5^10)`
`=(1/5)^10`
a)\({\left[ {{{\left( { - \frac{1}{6}} \right)}^3}} \right]^4}\) (với \(a = - \frac{1}{6}\))
\(=(- \frac{1}{6})^{3. 4}=(- \frac{1}{6})^{12}\)
b)\({\left[ {{{\left( { - 0,2} \right)}^4}} \right]^5}\) (với \(a = - 0,2\))
\(=(-0,2)^{4.5}=(-0,2)^{20}\)
a) \({\left( {\frac{8}{9}} \right)^3} \cdot \frac{4}{3} \cdot \frac{2}{3} = {\left( {\frac{8}{9}} \right)^3}.\frac{8}{9} = {\left( {\frac{8}{9}} \right)^{3+1}}={\left( {\frac{8}{9}} \right)^4}\)
b) \({\left( {\frac{1}{4}} \right)^7} \cdot 0,25 = {\left( {0,25} \right)^7}.0,25 ={\left( {0,25} \right)^{7+1}}= {\left( {0,25} \right)^8}\)
c) \({( - 0,125)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^6}:\frac{{ - 1}}{8} = {\left( {\frac{{ - 1}}{8}} \right)^{6-1}}= {\left( {\frac{{ - 1}}{8}} \right)^5}\)
d) \({\left[ {{{\left( {\frac{{ - 3}}{2}} \right)}^3}} \right]^2} = {\left( {\frac{{ - 3}}{2}} \right)^{3.2}} = {\left( {\frac{{ - 3}}{2}} \right)^6}\)
a) \({2^m}{.2^n}=\underbrace {2.2 \ldots .2}_{m{\rm{ }}}{\rm{ }}.\underbrace {2.2 \ldots .2}_{n{\rm{ }}}{\rm{ }}\) = 2m+n
b) \({3^m}:{3^n}=(\underbrace {3.3 \ldots .3}_{m{\rm{ }}}{\rm{ }}):(\underbrace {3.3 \ldots .3}_{n{\rm{ }}}{\rm{ }})\) = 3m-n với \(m \ge n\)
\(3,=\left(\dfrac{13}{25}-\dfrac{38}{25}\right)+\left(\dfrac{14}{9}-\dfrac{5}{9}\right)=-1+1=0\\ 4,=\left(\dfrac{4}{9}\right)^5\cdot\left(\dfrac{9}{49}\right)^5=\left(\dfrac{4}{9}\cdot\dfrac{9}{49}\right)^5=\left(\dfrac{4}{49}\right)^5\\ 5,\Rightarrow\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{x+y}{5+3}=\dfrac{2}{2}=\dfrac{x+y}{8}\Rightarrow x+y=8\\ 6,\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\Rightarrow2\text{ giá trị}\\ 7,=\dfrac{3^{10}\cdot2^{30}}{2^9\cdot3^9\cdot2^{20}}=2\cdot3=6\)
a: \(=\dfrac{3^3\cdot2^6}{3^{-4}\cdot2^6}=3^7\)
b: \(=\left(\dfrac{3}{7}\right)^5\cdot\left(\dfrac{3}{7}\right)\cdot\dfrac{5^6}{3^6}:\left(\dfrac{625}{343}\right)^2\)
\(=\dfrac{3^6}{7^6}\cdot\dfrac{5^6}{3^6}:\dfrac{5^8}{7^6}\)
\(=\dfrac{1}{5^2}\)
c: \(=5^{4+3}\cdot\left(\dfrac{5}{2}\right)^{-5}\cdot\dfrac{1}{25}\)
\(=5^5\cdot\left(\dfrac{2}{5}\right)^5=2^5\)
\(a)\left[\left(1\dfrac{2}{3}\right)^2\right]^3\\ =\left(1\dfrac{2}{3}\right)^{2\cdot3}\\ =\left(\dfrac{5}{3}\right)^6\\ b)\left(-0,18^3\right)^7\\=\left(-0,18\right)^{3\cdot7}\\ =\left(-0,18\right)^{21}\\ c)\left(\dfrac{1}{3^2}\right)^6\\ =\left[\left(\dfrac{1}{3}\right)^2\right]^6\\ =\left(\dfrac{1}{3}\right)^{2\cdot6}\\ =\left(\dfrac{1}{3}\right)^{12}\)