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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, \(\left(4.2\right)^5:\left(2^3.\dfrac{1}{16}\right)=8^5:\left(2^3.\dfrac{1^4}{2^4}\right)=\left(2^3\right)^5:\dfrac{2^3.1^4}{2^4}=2^{15}:\dfrac{1}{2}=2^{15}.2=2^{16}\)
\(b,\dfrac{2^2.4.32}{2^2.2^5}=\dfrac{2^2.2^4.2^5}{2^2.2^5}=2^4=16\)
\(a,\dfrac{\left(4.2\right)^5}{2^3.\dfrac{1}{16}}=\dfrac{\left(2^3\right)^5}{2^3.2^{-4}}=\dfrac{2^{15}}{2^{-1}}=2^{16}\)
b,\(\dfrac{2^2.4.32}{2^2.2^5}=\dfrac{2^2.2^2.2^5}{2^2.2^5}=2^2=4\)
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: \(=\dfrac{1}{27}\cdot3^4=3^1\)
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
a) Khoảng \(\left( { - 2;3} \right)\)
b) Đoạn \(\left[ {1;10} \right]\)
c) Nửa khoảng \(\left( {\left. { - 5;\sqrt 3 } \right]} \right.\)
d) Nửa khoảng \(\left. {\left[ {\pi ;4} \right.} \right)\)
e) Khoảng \(\left( { - \infty ;\frac{1}{4}} \right)\)
g) Nửa khoảng \(\left[ {\left. {\frac{\pi }{2}; + \infty } \right)} \right.\)
a: \(=2^2\cdot9\cdot\dfrac{1}{6\cdot9}\cdot\dfrac{4^2}{9^2}=\dfrac{2^2}{6}\cdot\dfrac{2^4}{3^4}=\dfrac{2^6}{2\cdot3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\left(\dfrac{1}{2}\right)^3\cdot2^3\cdot\left(\dfrac{1}{2}\right)^2}{\left(-8\right)^2\cdot16}\cdot2^6=\dfrac{\dfrac{1}{2^2}}{64\cdot16}\cdot64=\dfrac{1}{4}:16=\dfrac{1}{64}=\left(\dfrac{1}{8}\right)^2\)
a: \(=2^2\cdot9\cdot\dfrac{1}{3^3\cdot2}\cdot\dfrac{2^4}{3^4}=\dfrac{2^4\cdot2^2}{2}\cdot\dfrac{9}{3^3\cdot3^4}=\dfrac{2^5}{3^5}=\left(\dfrac{2}{3}\right)^5\)
b: \(=2^8\cdot\dfrac{3^4}{2^4}=3^4\cdot2^4=6^4\)
c: \(=\dfrac{\dfrac{1}{2^3}\cdot\dfrac{1}{2^2}\cdot8}{\left(-8\right)^2\cdot2^4}\cdot2^6=\dfrac{1}{2^2}\cdot2^6:2^{10}=\dfrac{2^4}{2^{10}}=\dfrac{1}{2^6}=\left(\dfrac{1}{8}\right)^2\)
a) \(3^3\)
b)\(2^8\)
c) \(2^7\)
d) \(3^1\)
a) 9.33.\(\dfrac{1}{81}\) .32 = 32. 33.\(\dfrac{1}{3^4}\) . 32 = 33
b) 4. 25: \(\) (23.\(\dfrac{1}{16}\))= 22. 25: 23. \(\dfrac{1}{2^4}\) = 27: \(\dfrac{1}{2}\) = 27. 2= 28
c) 32. 25. \(\left(\dfrac{2}{3}\right)^2\) = 32. 25. \(\dfrac{2^2}{3^2}\) = 25. 22 = 27
d) \(\left(\dfrac{1}{3}\right)^2\) .\(\dfrac{1}{3}\) . 92 = \(\dfrac{1}{9}.\dfrac{1}{3}\). 92 = \(\dfrac{9}{3}\) = 31