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3x + 3x+1 + 3x+2 = 117
=> 3x + 3x.3 + 3x.32 = 117
=>3x(1+3+32) = 117
=>3x.13 = 117
=>3x = 9
=>3x=32
=>x=3
3x + 3x+1 + 3x+2 = 117
=> 3x + 31 + 32 + 3x + 3x = 117
=> 3x . ( 31 + 32 + 1 ) = 117
=> 3x . 13 = 117
=> 3x = 117 : 13
=> 3x = 9
=> 3x = 32
=> x = 2
1.b) \(\left(\left|x\right|-3\right)\left(x^2+4\right)< 0\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|-3\\x^2+4\end{cases}}\) trái dấu
\(TH1:\hept{\begin{cases}\left|x\right|-3< 0\\x^2+4>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|< 3\\x^2>-4\end{cases}}\Leftrightarrow x\in\left\{0;\pm1;\pm2\right\}\)
\(TH1:\hept{\begin{cases}\left|x\right|-3>0\\x^2+4< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\left|x\right|>3\\x^2< -4\end{cases}}\Leftrightarrow x\in\left\{\varnothing\right\}\)
Vậy \(x\in\left\{0;\pm1;\pm2\right\}\)
Ta có : \(\frac{\left(4^x\right)^2}{2^x}=8\)
\(\Rightarrow4^{2x}=8.2^x\)
\(\Rightarrow4^{2x}=2^3.2^x\)
\(\Rightarrow\left(2^2\right)^{2x}=2^{x+3}\)
\(\Rightarrow2^{4x}=2^{x+3}\)
=> 4x = x + 3
=> 3x = 3
=> x = 1
Vậy x = 1.
1) \(\left|x\right|< 4\Leftrightarrow-4< x< 4\)
2) \(\left|x+21\right|>7\Leftrightarrow\orbr{\begin{cases}x+21>7\\x+21< -7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>-14\\x< -28\end{cases}}\)
3) \(\left|x-1\right|< 3\Leftrightarrow-3< x-1< 3\Leftrightarrow-2< x< 4\)
4) \(\left|x+1\right|>2\Leftrightarrow\orbr{\begin{cases}x+1>2\\x+1< -2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x>1\\x< -3\end{cases}}\)
\(\left|x+\frac{1}{2}\right|+\left|3-y\right|=0\)
Vì \(\hept{\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|3-y\right|\ge0\end{cases}}\Rightarrow\)\(\left|x+\frac{1}{2}\right|+\left|3-y\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left|x+\frac{1}{2}\right|=0\\\left|3-y\right|=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{-1}{2}\\y=3\end{cases}}\)
a)\(\frac{1}{4}-\frac{1}{3}x=\frac{2}{5}-\frac{3}{2}x\)
\(\Leftrightarrow\)\(\frac{15-20x}{60}=\frac{24-90x}{60}\)
\(\Leftrightarrow15-20x=24-90x\)
\(\Leftrightarrow-20x+90x=24-15\)
\(\Leftrightarrow70x=9\)
\(\Leftrightarrow x=\frac{9}{70}\)
c) (1/2-1/6)*3^x+4-4*3^x=3^16-4*3^13
=1/3*3^x*3^4-4*3^x=3^13*3^3-4*3^13
=27*3^x-4*3^x=3^13*(27-4)
=3^x*(27-4)=3^13*(27-4)
=>x=13
\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Rightarrow3^x+3^x.3+3^x.9=117\)
\(\Rightarrow3^x.\left(1+3+9\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=117:13\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Rightarrow3^x+3^x.3+3^x.9=117\)
\(\Rightarrow3^x.\left(1+3+9\right)=117\)
\(\Rightarrow3^x=117:13\)
\(\Rightarrow3^x=9\)
\(\Leftrightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)