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3\(^x\) = \(\dfrac{9^4}{81^3}\)
3\(^x\) = \(\dfrac{\left(3^2\right)^4}{\left(3^4\right)^3}\)
3\(^x\) = \(\dfrac{3^8}{3^{12}}\)
3\(^x\) = 3-4
\(x\) = -4
Vậy \(x\) = -4
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`#3107.101107`
\(x^4=\dfrac{1}{81}\\ \Rightarrow x^4=\left(\dfrac{1}{3}\right)^4\\ \Rightarrow x=\dfrac{1}{3}\)
Vậy, `x =`\(\dfrac{1}{3}.\)
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a)\(\left(\frac{3}{5}\right)^5\times x=\left(\frac{3}{7}\right)^7\)
\(\Leftrightarrow\frac{3^5}{5^5}\times x=\frac{3^7}{7^7}\)
\(\Leftrightarrow x=\frac{3^7}{7^7}:\frac{3^5}{5^5}\)
\(\Leftrightarrow x=\frac{3^7\times5^5}{7^7\times3^5}\)
\(\Leftrightarrow x=\frac{3^2\times5^5}{7^7}\)
b)\(\left(\frac{-1}{3}\right)^3\times x=\frac{1}{81}\)
\(\Leftrightarrow\frac{\left(-1\right)^3}{3^3}\times x=\frac{1}{3^4}\)
\(\Leftrightarrow x=\frac{1}{3^4}:\frac{-1}{3^3}\)
\(\Leftrightarrow x=\frac{1\times3^3}{3^4\times\left(-1\right)}\)
\(\Leftrightarrow x=\frac{1}{-3}\)
c)\(\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)
\(\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{3}+\frac{1}{2}\)
\(\Leftrightarrow x=\frac{5}{6}\)
d)\(\Leftrightarrow\left(x+\frac{1}{2}\right)^4=\left(\frac{2}{3}\right)^4\)
\(\Leftrightarrow x+\frac{1}{2}=\frac{2}{3}\)
\(\Leftrightarrow x=\frac{2}{3}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{6}\)
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3^2x+1=81
3^2x+1=3^4
x+1=3^4:3^2
x+1=3^2
x+1=9
x=9-1
x=8
vậy x=8 hoặc x=2^3
đúng thì nhớ k nhé
kb luon
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\(\frac{81}{3^{2x}+1}=3\)
=> 3.(32x + 1) = 81
=> 32x+1 = 27
=> 32x+1 = 33
=> 2x + 1 = 3
=> 2x = 2
=> x = 1
Đúng 100%
\(\Rightarrow3^{2x+1}=\frac{9^4}{9^2}\)
\(\Rightarrow3^{2x+1}=9^2\)
\(\Rightarrow2x+1=4\)
\(\Rightarrow2x=3\)
\(\Rightarrow x=\frac{3}{2}\)
\(3^{2x-1}=9^2\)
\(3^{2x-1}=3^4\)
\(\Rightarrow2x-1=4\)
\(2x=4+1\)