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3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹.(1 + 3 + 3²) = 39
3ˣ⁻¹.13 = 39
3ˣ⁻¹ = 39 : 13
3ˣ⁻¹ = 3
x - 1 = 1
x = 1 + 1
x = 2
\(3^{x-1}+3^x+3^{x+1}=39\)
\(=>3^x:3+3^x+3^x.3=39\)
\(=>3^x.\dfrac{1}{3}+3^x+3^x.3=39\)
\(=>3^x.\left(\dfrac{1}{3}+1+3\right)=39\)
\(=>3^x.\dfrac{13}{3}=39\)
\(=>3^x=39:\dfrac{13}{3}=39.\dfrac{3}{13}\)
\(=>3^x=9=3^2\)
\(=>x=2\)
\(3^{x-1}+3^x+3^{x+1}=39\)
\(3^{x-1}+3^{x-1}.3+9.3^{x-1}=39\)
\(13.3^{x-1}=39\)
\(3^{x-1}=39:13=3\)
\(x-1=1\)
\(x=2\)
Sửa đề: 3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹ + 3ˣ + 3ˣ⁺¹ = 39
3ˣ⁻¹.(1 + 3 + 3²) = 39
3ˣ⁻¹ . 13 = 39
3ˣ⁻¹ = 39 : 13
3ˣ⁻¹ = 3
x - 1 = 1
x = 1 + 1
x = 2
Ta có : \(\frac{3x}{2\times5}+\frac{3x}{5\times8}+\frac{3x}{8\times11}+\frac{3x}{11\times14}=\frac{1}{21}\)
\(\Rightarrow x\times\left(\frac{3}{2\times5}+\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}\right)=\frac{1}{21}\)
\(\Rightarrow x\times\left(\frac{1}{2\times5}+\frac{1}{5\times8}+\frac{1}{8\times11}+\frac{1}{11\times14}\right)=\frac{1}{21}\)
\(x\times\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)=\frac{1}{21}\)
\(x\times\left(\frac{1}{2}-\frac{1}{14}\right)\) \(=\frac{1}{21}\)
\(x\times\frac{3}{7}\) \(=\frac{1}{21}\)
\(x\) \(=\frac{1}{21}\div\frac{3}{7}=\frac{1}{21}\times\frac{7}{3}\)
\(\Rightarrow x=\frac{1}{9}\)
Ta có 3x/2.5+3x/5.8+3x/8.11+3x/11.14=1/21
=>x(3/2.5+3/5.8+3/8.11+3/11.14)=1/21
=>3x(1/2.5+1/5.8+1/8.11+1/11.14)=1/21
=>3x(1/2-1/14)=1/21
=>3x.3/7=1/21
=>3x=1/21:3/7
=>3x=1
=>x=1:3=1/3
a,(2x-1)3 =23+102 b,(3x+1)+(3x+3)+...+(3x+99)=2800
(2x-1)3 =125 3x+1+3x+3+...+3x+99=2800
(2x-1)3=53 ( 3x+3x+.....+3x )+(1+3+...+99)=2800
2x-1=5 gọi A=3x+3x+...+3x ; B=1+3+...+99
2x=5+1 số số hạng của B là : (99-1):2+1=50 ( bằng số số hạng của A)
2x=6 B = (99+1) x 50:2
=2500
x=6:2 ta có: 150x + 2500=2800
x=3 150x=2800-2500
vậy x=3 150x=300
x=300:150
x=2
vậy x=2
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
Ta có:
\(\frac{3x}{2.5}+\frac{3x}{5.8}+\frac{3x}{8.11}+\frac{3x}{11.14}\)
\(=x.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}\right)\)
\(=x.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\right)\)
\(=x.\left(\frac{1}{2}-\frac{1}{14}\right)\)
\(=x.\left(\frac{7}{14}-\frac{1}{14}\right)=x.\frac{3}{7}=\frac{1}{21}\)
\(\Rightarrow x=\frac{1}{21}:\frac{3}{7}=\frac{1}{21}.\frac{7}{3}=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
\(3^x+3^{x+1}+3^{x+2}=117\)
\(3^x+3^x.3+3^x.3^2=117\)
\(3^x\left(1+3+3^2\right)=117\)
\(3^x.13=117\)
\(3^x=9\)
\(\Rightarrow x=2\)
\(\Leftrightarrow3^{x-1}\left(1+3+3^2\right)=39\\ \Leftrightarrow3^{x-1}\cdot13=39\\ \Leftrightarrow3^{x-1}=3=3^1\\ \Leftrightarrow x-1=1\Leftrightarrow x=2\)
\(\Leftrightarrow3^x\cdot\dfrac{13}{3}=39\)
\(\Leftrightarrow x=2\)
\(\Leftrightarrow3.3^x-\dfrac{3^x}{3}=72\)
\(\Leftrightarrow9.3^x-3^x=3.72\)
\(\Leftrightarrow8.3^x=3.8.3^2\)
\(\Leftrightarrow3^x=3^3\Rightarrow x=3\)
\(3^{x-1}+3^x+3^{x+1}-1=1052\)
\(3^x:3+3^x+3^x\cdot3=1053\)
\(3^x\left(\frac{1}{3}+1+3\right)=1053\)
\(3^x\cdot\frac{13}{3}=1053\)
\(3^x=243\)
\(x=5\)
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