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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 \(x=2\)
a,(=)\(3^{x+1}.\left(3+4\right)=7.3^6\)
(=)\(3^{x+1}=3^6\)
=>x+1=6(=)x=5
b
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
a: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
\(\Leftrightarrow2^x=2^{10}\cdot5:\dfrac{5}{2}=2^{10}\cdot5\cdot\dfrac{2}{5}=2^{11}\)
=>x=11
b: \(\Leftrightarrow3^x\cdot\dfrac{1}{3}+3^x\cdot9=3^{13}\cdot28\)
\(\Leftrightarrow3^x=3^{13}\cdot28:\dfrac{28}{3}=3^{14}\)
hay x=14
1: \(5\cdot3^x=5\cdot3^4\)
nên \(3^x=3^4\)
hay x=4
2: \(7\cdot4^x=7\cdot4^3\)
nên \(4^x=4^3\)
hay x=3
3: \(8\cdot7^x=8\cdot7^6\)
nên \(7^x=7^6\)
hay x=6
2: =>2x-1/4=5/6-1/2x
=>5/2x=5/6+1/4=13/12
=>x=13/30
3: =>3x-5/6=2/3-1/2x
=>3,5x=2/3+5/6=4/6+5/6=9/6=3,2
hay x=32/35
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(1+3+9)=117
3x.13=117
3 x=9
=>x=2