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1. Tim x
a. ( \(\frac{8}{27}\))x = ( \(\frac{2}{3}\))72
b. \(\frac{1}{3}\) . 3x = 7 . 32 . 92 - 2 . 3x
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a. \(\left(\frac{8}{27}\right)^x=\left(\frac{2}{3}\right)^{72}\)
\(\left(\frac{2}{3}\right)^{3x}=\left(\frac{2}{3}\right)^{72}\)
\(\Rightarrow3x=72\Rightarrow x=24\)
Vậy x = 24
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\(\frac{1}{81}.27^x=3^x\Rightarrow\frac{1^2}{9^2}=\frac{3^x}{27^x}\)\(\Rightarrow\left(\frac{1}{9}\right)^2 =\left(\frac{1}{9}\right)^x\)\(\Rightarrow X=2\)
=>\(\frac{1}{3^4}\cdot\left(3^3\right)^x=3^x\)
=>\(\frac{3^{3x}}{3^4}=3^x\)
=>\(3x-4=x\)
=>\(3x=x+4\)
=>\(2x=4=>x=2\)
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a) 3x = 81 = 34 => x = 4
b) (x+1)3 = 27 = 33
=> x + 1 = 3 => x = 2
c) 7x+2 = 27. x thuộc N => x + 2 thuộc N.
Mà 71 = 7 < 7x+2 = 27 < 72 = 49; nghĩa là 1 < x + 2 < 2
Do đó ko tồn tại x thỏa mãn
a) 3x = 81 = 34
=> x = 4
b) (x+1)3 = 27 = 33
=> x + 1 = 3
=> x = 2
c) 7x+2 = 27.
Vì \(x\in N\) nên \(x\in\phi\)
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2|x - 5| = 8
\(\Rightarrow\) |x - 5| = 4
\(\Rightarrow\) |x - 5| = \(\left\{{}\begin{matrix}4\\-4\end{matrix}\right.\)
\(\Rightarrow\) x = \(\left\{{}\begin{matrix}9\\1\end{matrix}\right.\)
\(3^{x+1}=27\)
\(3^{x+1}=3^3\)
\(\Rightarrow x+1=3\)
\(x=3-1\)
\(x=2\)
Vậy x = 2.
\(#Paciupibijd\)
\(3^{x+1}=27\)
\(\Rightarrow3^{x+1}=3^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=3-1\)
\(\Rightarrow x=2\)