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Em kiểm tra lại đề bài, \(3^x+3^{x+2}+3^{x+2}\) hay \(3^x+3^{x+1}+3^{x+2}\)
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
Ta có: \(3x+3x+1+3x+2+3x+3=29160\)
\(\Leftrightarrow12x=29154\)
hay x=2429,5
\(3x+3x+1+3x+2+3x+3=29160\Rightarrow3x+3x+3x+3x=29160-1-2-3\)
\(\Rightarrow12x=29154\Rightarrow x=\dfrac{4859}{2}=2429,5\)
\(=3^{x+1}\left(1+3+3^2\right)+...+3^{x+10}\left(1+3+3^2\right)=\)
\(=3^x.3.13+...+3^{x+9}.3.13=\)
\(39\left(3^x+...+3^{x+9}\right)⋮39\)
a) Ta có 2x + 2x + 1 + 2x + 2 = 56
⇒ 2x ( 1 + 21 + 22 ) = 56
⇒ 2x . 7 = 56
⇒ 2x = 56 : 7 = 8 = 23
Vậy x = 3
b) Ta có 3x + 3x + 2 + 3x + 3 = 111
⇒ 3x ( 1 + 32 + 33 ) = 111
⇒ 3x . 37 = 111
⇒ 3x = 111 : 37 = 3 = 31
Vậy x = 1
\(\left(3x-1\right)^3=25\left(3x-1\right)\\ \Leftrightarrow\left(3x-1\right)^2=25\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\\ \left(3x-14\right)^3=2^5\cdot5^2+200\\ \Leftrightarrow\left(3x-14\right)^3=1000=10^3\\ \Leftrightarrow3x-14=10\Leftrightarrow x=8\)
\(\left(3x-1\right)^3=25\left(3x-1\right)\)
\(\Rightarrow\left(3x-1\right)\left(9x^2-6x+1-25\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(9x^2-6x-24\right)=0\)
\(\Rightarrow3\left(3x-1\right)\left(x-2\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
\(\left(3x-14\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(3x-14\right)^3=1000\)
\(\Rightarrow3x-14=10\Rightarrow3x=24\Rightarrow x=8\)
1, 4[3x + 4] = 12x + 16
2, 4/3[5/3x + 6] = 20/9x + 12
3, [4/3x + 7]. 5 + 2 = 20/3x + 35 + 2 = 20/3x + 37
4, [3/7x - 6][-7] + 3x = -3x + 42 + 3x = 42
Ta có: \(3^x+3^{x+2}+3^{x+3}=3^5\cdot37\)
\(\Leftrightarrow3^x\cdot\left(1+9+27\right)=3^5\cdot37\)
hay x=5