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a) \(9\cdot3^3\cdot\frac{1}{81}\cdot3^2\)
\(=\frac{3^2\cdot3^3\cdot3^2}{3^4}\)
\(=3^3=27\)
b) \(4\cdot2^5:\left(2^3\cdot\frac{1}{16}\right)\)
\(=\frac{2^2\cdot2^2\cdot2^4}{2^3}\)
\(=2^5=32\)
c) \(3^2\cdot2^5\cdot\left(\frac{2}{3}\right)^2\)
\(=\frac{3^2\cdot2^5\cdot2^4}{3^2}\)
\(=2^9=512\)
d) \(\left(\frac{1}{3}\right)^2\cdot\frac{1}{3}\cdot9^2\)
\(=\frac{1^2\cdot1\cdot3^4}{3^2}\)
\(=3^2=9\)
a: \(=3^2\cdot3^3\cdot3^{-4}\cdot3^2=3^{2+3-4+2}=3^3\)
b: \(=2^2\cdot2^5:\left(2^3\cdot\dfrac{1}{2^4}\right)=2^7:\dfrac{1}{2}=2^8\)
c: \(=9\cdot32\cdot\dfrac{4}{9}=128=2^7\)
d: \(=\dfrac{1}{27}\cdot3^4=3^1\)
B1:
Vì \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y-\frac{1}{3}\right|\ge0\\\left|4z+5\right|\ge0\end{cases}\left(\forall x,y,z\right)}\Rightarrow\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\ge0\left(\forall x,y,z\right)\)
Mà theo đề bài, \(\left|x-\frac{1}{2}\right|+\left|2y-\frac{1}{3}\right|+\left|4z+5\right|\le0\) nên dấu "=" xảy ra khi:
\(\left|x-\frac{1}{2}\right|=\left|2y-\frac{1}{3}\right|=\left|4z+5\right|=0\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{6}\\z=-\frac{5}{4}\end{cases}}\)
a. \(25.5^3.\frac{1}{625}.5^2=5^2.5^3.\frac{1}{5^4}.5^2=\frac{5^7}{5^4}=5^3\)
b. \(4.32:\left(2^3.\frac{1}{16}\right)=2^2.2^5:2^3:\frac{1}{2^4}=\frac{2^4}{2^4}=1\)
c. \(5^2.3^5.\left(\frac{3}{5}\right)^2=5^2.3^5.3^2.\frac{1}{5^2}==\frac{5^2}{5^2}.3^7=3^7\)
d. \(\left(\frac{1}{7}\right)^2.\frac{1}{7}.49^2=\frac{1}{7^3}.7^4=\frac{7^4}{7^3}=7\)
a) \(3^2.2^5.\left(\frac{2}{3}\right)^2=2^5.\frac{2^2.3^2}{3^2}=2^5.2^2=2^7\)
b) \(\left(\frac{1}{3}\right)^2.\frac{1}{3}.9^2=\frac{1^2.1}{3^2.3}.\left(3^2\right)^2=\frac{1.3^{2.2}}{3^3}=\frac{3^4}{3^3}=3^1\)