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\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Rightarrow3^x+3^x.3+3^x.9=117\)
\(\Rightarrow3^x.\left(1+3+9\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=117:13\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
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 \(x=2\)
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
=> 3^x . (1+3+3^2) = 117
=> 3^x . 13 = 117
=> 3^x = 117 : 13 = 9
=> 3^x = 3^2
=> x = 2
Vậy x = 2
Tk mk nha
3x + 3x+1 +3x+2 = 117
=> 3x + 3x . 3 + 3x . 9 = 117
=> 3x ( 1 + 3 + 9) = 117
=> 3x . 13 = 117
=> 3x = 9
=> x = 2
3^x + 3^x+1 + 3^x+2=117
3^x.(1+3+3^2)=117
3^x.13=117
3^x=117:13
3^x=9
Ta có:9=3^2
Vậy x=2
\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Rightarrow3^x+3^x.3+3^x.3^2=117\)
\(\Rightarrow3^x.\left(1+3+3^2\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=9\)
\(\Rightarrow3^x=3^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
\(\Rightarrow3^x\left(1+3+3^2\right)=117\)
\(\Rightarrow3^x.13=117\)
\(\Rightarrow3^x=9\)
=>x = 2
Vây x = 2
a) \(\frac{5}{6}\)x - \(\frac{3}{8}\)x - 10 = 12
=> \(\left(\frac{5}{6}-\frac{3}{8}\right)\)x = 12 + 10
=> \(\frac{11}{24}\)x = 22
=> x = 22 : \(\frac{11}{24}\)
=> x = 48
Vậy x = 48.
b) (\(\left|x\right|\) - \(\frac{1}{8}\)) . \(\left(-\frac{1}{8}\right)^5\) = \(\left(-\frac{1}{8}\right)^7\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^7\) : \(\left(-\frac{1}{8}\right)^5\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\left(-\frac{1}{8}\right)^{7-5}\)
=> \(\left|x\right|\) - \(\frac{1}{8}\) = \(\frac{1}{64}\)
=> \(\left|x\right|\) = \(\frac{1}{64}\) + \(\frac{1}{8}\)
=> \(\left|x\right|\) = \(\frac{9}{64}\)
=> x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)
Vậy x = \(\frac{9}{64}\) hoặc x = \(\frac{-9}{64}\)
hên quá làm đúng hì hì, cảm ơn nhen, hết sợ bị sai ồi
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 \(x\in\left\{0;\pm1;\pm2\right\}\)
a)Ta có:
\(3^x-3^{x-3}=-234\)
\(\Rightarrow3^x-3^x\cdot3^3=-234\)
\(\Rightarrow3^x\cdot\left(1-3^3\right)=-234\)
\(\Rightarrow3^x\cdot\left(-26\right)=-234\)
\(\Rightarrow3^x=9\)
\(\Rightarrow x=2\)
Vậy x=2
\(\Rightarrow3^x=3^2\)
b) Ta có:
\(2^{x+1}\cdot3^x-6^x=216\)
\(\Rightarrow2^x\cdot2\cdot3^x-2^x\cdot3^x=216\)
\(\Rightarrow\left(2^x\cdot3^x\right)\cdot\left(2-1\right)=216\)
\(\Rightarrow6^x\cdot1=216\)
\(\Rightarrow6^x=6^3\)
\(\Rightarrow x=3\)
Vậy 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
nguyenmanhtrung hám **** ko lm được ý mà