tìm x biết
3x + 3x+1 +3x +2 117
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\(3x+3x+1+3x+2=117\)
\(\Rightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)
\(\Rightarrow9x+3=117\)
\(\Rightarrow9x=117-3\)
\(\Rightarrow9x=114\)
\(\Rightarrow x=114:9\)
\(\Rightarrow x=\frac{38}{3}\)
Vậy \(x=\frac{38}{3}\)
P/s : Đúng nha
~ Ủng hộ nhé
a)
\(3^x+3^{x+1}+3^{x+2}=117\\ \Leftrightarrow3^x+3.3^x+9.3^x=117\\ 13.3^x=117\\ \Leftrightarrow3^x=9\\ \Leftrightarrow3^x=3^2\\ \Leftrightarrow x=2\)
b)
\(3+4\left(x-10\right)=3^2+6\\ \Leftrightarrow3+4\left(x-10\right)=15\\ \Leftrightarrow4\left(x-10\right)=12\\ \Leftrightarrow x-10=3\\ \Leftrightarrow x=13\)
a) \(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x.\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(x=2\)
b) \(3+4\left(x-10\right)=3^2+6\)
\(3+4x-40=9+6\)
\(4x=15+40-3\)
\(4x=52\)
\(x=13\)
\(3x+3x+1+3x+2=117\)
\(\Leftrightarrow\left(3x+3x+3x\right)+\left(1+2\right)=117\)
\(\Leftrightarrow9x+3=117\)\(\Rightarrow9x=114\Rightarrow x=\frac{114}{9}\)
\(\text{Vậy x=}\frac{114}{9}\)
\(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(\Rightarrow x=2\)
Bài 1:
Thay \(x=\frac{4}{3};y=-1\)vào biểu thức A, ta được:
\(A=\frac{4}{3}\cdot\left[3\cdot\frac{4}{3}-\left(-1\right)\right]-\left(3\cdot\frac{4}{3}+1\right)\left(-1\right)\)
\(A=\frac{20}{3}+5=\frac{35}{3}\)
Vậy khi \(x=\frac{4}{3};y=-1\)thì A=\(\frac{35}{3}\)
\(B=3\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{8}{39}\)
\(B=\frac{352}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot\frac{713}{119}-\frac{8}{39}=-\frac{412}{1071}\)
\(3x+3x+1+3x+2=117\)
\(\Rightarrow3x+3x+3x=117-1-2\)
\(\Rightarrow3x+3x+3x=114\)
\(\Rightarrow x.\left(3+3+3\right)=114\)
\(\Rightarrow x.9=114\)
\(\Rightarrow x=\dfrac{38}{3}\)
Vậy \(x=\dfrac{38}{3}\)
=> 3x+3x+3x+1+2=117
=>9x+3=117
=>9x=117-3=114
=> x=\(\dfrac{114}{9}\)
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
Ta sẽ đưa các tích về 1 dãy tỉ số
\(3x=5y\Leftrightarrow\frac{x}{5}=\frac{y}{3}\Leftrightarrow\frac{x}{15}=\frac{y}{9},7y=9z\Leftrightarrow\frac{y}{9}=\frac{z}{7}\Rightarrow\frac{x}{15}=\frac{y}{9}=\frac{z}{7},x-y+z=117\left(gt\right)\)
Áp dụng tính chất dãy tỉ số bằng nhau cho dãy tỉ số trên ta được
\(\frac{x}{15}=\frac{y}{9}=\frac{z}{7}=\frac{x-y+z}{15-9+7}=\frac{117}{13}=9\Rightarrow x=15.9=135,y=9.9=81,z=7.9=63\)
Vậy \(x=135,y=81,z=63\)
Ta có: \(3x=5y=\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x}{15}=\frac{y}{9}\)
\(7y=9z=\frac{y}{9}=\frac{z}{7}\)
\(\Rightarrow\frac{x}{15}=\frac{y}{9}=\frac{z}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{15}=\frac{y}{9}=\frac{z}{7}=\frac{x-y+z}{15-9+7}=\frac{117}{13}=9\)
\(\Rightarrow\frac{x}{15}=9\Rightarrow x=9\cdot15=135\)
\(\frac{y}{9}=9\Rightarrow y=9\cdot9=81\)
\(\frac{z}{7}=9\Rightarrow z=9\cdot7=63\)
Vậy x=135, y=81 và z=63
a: Ta có: \(3x\left(3x-1\right)-\left(3x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow9x^2-3x-9x^2+1=0\)
\(\Leftrightarrow3x=1\)
hay \(x=\dfrac{1}{3}\)
b: Ta có: \(x^2-5x+25-5x=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
hay x=5
\(3^x+3^{x+1}+3^{x+2}=117\)
\(\Leftrightarrow3^x+3^x.3^1+3^x.3^2=117\)
\(\Leftrightarrow3^x\left(1+3+3^2\right)=117\)
\(\Leftrightarrow3^x.13=117\)
\(\Leftrightarrow3^x=9=3^2\)
\(\Leftrightarrow x=2\)
3^x+3^x+1+3 ^x+2=117
=> 3^x .(1+3+3^2 )=117
=> 3^x .(1+3+9)=117
=> 3^x .13=117
=> 3^x=117:13
=> 3^x=9
=> 3^x=3^2
Vậy x=2