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3x + 3x + 1 + 3x + 2 + 3x + 3 = 29 160
=> 3x . ( 1 + 3 + 32 + 33 ) = 29 160
=> 3x . 40 = 29 160
=> 3x = 729
=> 3x = 36
=> x = 6
(2x - 19)2019= (2x - 19)3
=> ( 2x - 19 )2019 - ( 2x - 19 )3 = 0
=> ( 2x - 19 )3 . [ ( 2x - 19 )2016 - 1 ] = 0
\(\Rightarrow\orbr{\begin{cases}\left(2x-19\right)^3=0\\\left(2x-19\right)^{2016}-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-19=0\\\left(2x-19\right)^{2016}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x=19\\2x-19\in\left\{1;-1\right\}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{19}{2}\\2x\in\left\{20;18\right\}\Rightarrow x∈\left\{10;9\right\}\end{cases}}\)
Tìm x\(\in\)n
a,x3-23=25-(316:314+28:216)
b,5x-2-32=24-(68:66-62)
c,(x2-1)4=81
d,3x+42=196:(193.192)-3.1
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a) \(3^{x+1}.15=135\)
\(\Rightarrow3^{x+1}=9\)
\(\Rightarrow3^{x+1}=3^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
b) \(x+2x+2^2x+....+2^{2016}x=2^{2017}-1\\ \Rightarrow x\left(2+2^2+...+2^{2016}\right)=2^{2017}-1\\ \Rightarrow x\left(2^{2017}-2\right)=2^{2017}-1\)
c) \(x\left(x-1\right)+\left(x-1\right)^2=0\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+\left(x-1\right)\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\begin{cases}x-1=0\\2x-1=0\end{cases}\)
d) \(2^2.2^5\le2^{x-5}\le2^{10}\\ \Rightarrow2^7\le2^{x-5}\le2^{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 135 - 5(x + 4) = 35
<=> 135 - 5x - 20 = 35
<=> 115 - 5x = 35
<=> 5x = 115 - 35
<=> 5x = 80
<=> x = 16
b) 25 + 3 (x - 8) = 106
=> 3 ( x - 8) = 106 - 25
=> 3 ( x - 8) = 81
=> ( x - 8) = 81: 3
=> x - 8 = 27
=> x = 27 + 8
=> x = .......