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2x+ 2x+1+ 2x+2 + 2x+3 = 2x+ 2x.2 + 2x.22 + 2x.23 = 2x(1+ 2 + 22 + 23) = 2x.15 = 120 => 2x =120:15 = 8 => x = 3
2x + 2x+1 +2x+2 + 2x+3 = 120
Suy ra 2x .1 + 2x . 2 + 2x . 22 + 2x . 23 = 120
Suy ra 2x .( 1 + 2 + 4 + 8 ) = 120
Suy ra 2x . 15 = 120
Suy ra 2x = 120 : 15
Suy ra 2x = 8
Suy ra 2x = 23
Do đó x = 3
Vậy x = 3
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
\(\Rightarrow2^x+2^x.2+2^x.2^2+2^x.2^3=120\)
\(\Rightarrow2^x+2^x.2+2^x.4+2^x.8=120\)
\(\Rightarrow2^x\left(1+2+4+8\right)=120\)
\(\Rightarrow2^x.15=120\)
\(\Rightarrow2^x=120:15\)
\(\Rightarrow2^x=8\)
\(\Rightarrow2^x=2^3\)
\(\Rightarrow x=3\)
a)
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\\ \Leftrightarrow2^x.1+2^x.2+2^x.2^2+2^x.2^3=120\\ \Leftrightarrow2^x\left(1+2+2^2+2^3\right)=120\\ \Leftrightarrow2^x=8=2^3\\ \Rightarrow x=3\)
b)
\(\dfrac{x+4}{2011}+\dfrac{x+3}{2012}=\dfrac{x+2}{2013}+\dfrac{x+1}{2014}\\ \Leftrightarrow\dfrac{x+4}{2011}+1+\dfrac{x+3}{2012}+1=\dfrac{x+2}{2013}+1+\dfrac{x+1}{2014}+1\\ \Leftrightarrow\dfrac{x+2015}{2011}+\dfrac{x+2015}{2012}=\dfrac{x+2015}{2013}+\dfrac{x+2015}{2014}\\ \Leftrightarrow\left(x+2015\right).\dfrac{1}{2011}+\left(x+2015\right).\dfrac{1}{2012}-\left(x+2015\right).\dfrac{1}{2013}-\left(x+2015\right).\dfrac{1}{2014}=0\\ \Leftrightarrow\left(x+2015\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2013}-\dfrac{1}{2014}\right)=0\\ \Rightarrow x+2015=0\Leftrightarrow x=-2015\)
`#3107.101107`
a)
\(27< 3^x< 243\\ \Rightarrow3^3< 3^x< 3^5\\ \Rightarrow3< x< 5\\ \Rightarrow x=4\)
Vậy, `x = 4`
b)
\(2^x+2^{x+1}+2^{x+2}=56?\\ \Rightarrow2^x+2^x\cdot2+2^x\cdot4=56\\ \Rightarrow2^x\cdot\left(1+2+4\right)=56\\ \Rightarrow2^x\cdot7=56\\ \Rightarrow2^x=8\\ \Rightarrow2^x=2^3\\ \Rightarrow x=3\)
Vậy, `x = 3`
c)
\(3^x+3^{x+2}=810\\ \Rightarrow3^x+3^x\cdot9=810\\ \Rightarrow3^x\cdot\left(1+9\right)=810\\ \Rightarrow3^x\cdot10=810\\ \Rightarrow3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
Vậy, `x = 4.`
a) \(27< 3^x< 243\)
\(\Rightarrow3^3< 3^x< 3^5\)
\(\Rightarrow3< x< 5\)
c) \(3^x+3^{x+2}=810\)
\(\Rightarrow3^x\left(1+3^2\right)=810\)
\(\Rightarrow3^x.10=810\)
\(\Rightarrow3^x=810:10\)
\(\Rightarrow3^x=81\)
\(\Rightarrow3^x=3^4\)
\(\Rightarrow x=4\)
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
\(\Rightarrow2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=120\)
\(\Rightarrow\left(1+2+4+8\right)\cdot2^x=120\)
\(\Rightarrow15\cdot2^x=120\)
\(\Rightarrow2^x=\dfrac{120}{15}=8=2^3\)
\(\Rightarrow x=3\)
Vậy............
\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)
\(=2^x.1+2^x.2^1+2^x.2^2+2^x.2^3=120\)
\(=2^x.1+2^x.2+2^x.4+2^x.8=120\)
\(=2^x\left(1+2+4+8\right)=120\)
\(=2^x.15=120\)
\(2^x=120:15=8\)
\(2^x=2^3\Leftrightarrow x=3\)