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\(-\frac{32}{\left(-2\right)^n}=4\\ =>-\frac{32}{\left(-2\right)^n}=\frac{4}{1}\\ =>\left(-2\right)^n=8\\ =>n=3\)
\(-\frac{32}{\left(-2\right)^n}=4\Rightarrow\frac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
=> (-2)5:(-2)n=(-2)2
=> 5-n=2
=> n=3
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\(\frac{2^{4-x}}{16^5}=32^6\)
=> \(\frac{2^{4-x}}{\left(2^4\right)^5}=\left(2^5\right)^6\)
=> \(\frac{2^{4-x}}{2^{20}}=2^{30}\)
=> \(2^{4-x}=2^{30}.2^{20}\)
=> \(2^{4-x}=2^{50}\)
=> 4 - x = 50
=> x = 4 - 50 = -46
\(\frac{3^{2x+3}}{9^3}=9^{14}\)
=> \(\frac{3^{2x+3}}{\left(3^2\right)^3}=\left(3^2\right)^{14}\)
=> \(\frac{3^{2x+3}}{3^6}=3^{28}\)
=> \(3^{2x+3}=3^{28}.3^6\)
=> \(3^{2x+3}=3^{34}\)
=> 2x + 3 = 34
=> 2x = 34 - 3
=> 2x = 31
=> x = 31/2
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a) \(\frac{-32}{\left(-2\right)^n}=4\)
\(\frac{\left(-2\right)^5}{\left(-2\right)^n}=4\)
\(\left(-2\right)^{5-n}=\left(-2\right)^2\)
=> 5-n = 2
n = 3
b) \(\frac{8}{2^n}=2\)
\(\frac{2^3}{2^n}=2\)
\(2^{3-n}=2^1\)
=> 3 -n = 1
n = 2
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)
=> 2n -1 = 3
2n = 4
n = 2
a) \(\frac{-32}{\left(-2\right)^n}=4\Leftrightarrow\left(-2\right)^n=\frac{-32}{4}\)
\(\left(-2\right)^n=-8\)Mà \(-8=2^{-3}\)
\(\Rightarrow x=-3\)
b) \(\frac{8}{2^n}=2\Leftrightarrow2^n=\frac{8}{2}\)
\(2^n=4\) Mà \(4=2^2\Rightarrow x=2\)
c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\Rightarrow\left(\frac{1}{2}\right)^{2n}:\frac{1}{2}=\frac{1}{8}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{8}\cdot\frac{1}{2}\)
\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{16}\Leftrightarrow\frac{1}{2^{2n}}=\frac{1}{16}\) mà\(16=2^4\)
\(2n=4\Rightarrow n=2\)
Vậy .........................
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a)\(\frac{21}{3^n}=3\)
\(\Leftrightarrow27:3^n=3\)
\(\Leftrightarrow3^n=27:3\)
\(\Leftrightarrow3^n=9\)
\(\Leftrightarrow3^n=3^2\)
\(\Rightarrow n=2\)
Vậy \(n=2\)
b) \(\frac{-32}{\left(-2\right)^n}=4\)
\(\Leftrightarrow\left(-32\right):\left(-2\right)^n=4\)
\(\Leftrightarrow\left(-2\right)^n=\left(-32\right):4\)
\(\Leftrightarrow\left(-2\right)^n=-8\)
\(\Leftrightarrow\left(-2\right)^n=\left(-2\right)^3\)
\(\Rightarrow n=3\)
Vậy \(n=3\)
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{n}{n+1}\)
\(=\frac{1}{n+1}\)
\(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+\frac{1}{4}.4.5:2+...+\frac{1}{20}.20.21:2\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{21}{2}\)
\(=\frac{2+3+4+5+...+21}{2}=115\)
(-2)^n.4=-32
(-2)^n.(-2)^2=(-2)^5
(-2)^n=-2^3
n=3
\(\frac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\left(-2\right)^n=\frac{-32\cdot1}{4}\)
\(\Rightarrow\left(-2\right)^n=-8\)
\(\Rightarrow\left(-2\right)^3=-8\)
\(\Rightarrow n=3\)