tìm x biết : 3x+1 - 3x =1458
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19 tháng 1 2022
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
13 tháng 8 2021
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
CC
1
ND
2
5 tháng 5 2021
Sửa đề: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
a) Ta có: 3x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=4153x1⋅5+3x5⋅9+3x9⋅13+...+3x81⋅85=415
⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415⇔3x4(41⋅5+45⋅9+49⋅13+...+481⋅85)=415
⇔x⋅34(1−15+15−19+19−113+...+181−185)=415⇔x⋅34(1−15+15−19+19−113+...+181−185)=415
⇔x⋅34(1−185)=415⇔x⋅34(1−185)=415
⇔x⋅6385=415⇔x⋅6385=415
hay x=68189x=68189
Vậy: x=68189
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{81\cdot85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{3}{4}\left(1-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
OD
1
5 tháng 5 2021
Sửa đề: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
a) Ta có: \(\dfrac{3x}{1\cdot5}+\dfrac{3x}{5\cdot9}+\dfrac{3x}{9\cdot13}+...+\dfrac{3x}{81\cdot85}=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{81}-\dfrac{1}{85}\right)=\dfrac{4}{15}\)
\(\Leftrightarrow\dfrac{3x}{4}\cdot\dfrac{84}{85}=\dfrac{4}{15}\)
\(\Leftrightarrow x\cdot\dfrac{63}{85}=\dfrac{4}{15}\)
hay \(x=\dfrac{68}{189}\)
Vậy: \(x=\dfrac{68}{189}\)
\(3^{x+1}-3^x=1458\)
\(\Leftrightarrow3^x.3-3^x=1458\)
\(3^x.\left(3-1\right)=1458\)
\(3^x.2=1458\)
\(3^x=1458:2\)
\(3^x=729\)
\(3^x=3^6\)
\(\Rightarrow x=6\)
3\(x+1\) - 3\(^x\) = 1458
3\(^x\).( 3 - 1) = 1458
3\(^x\). 2 = 1458
3\(^x\) = 1458: 2
3\(^x\) = 729
3\(^x\) = 36
\(x\) = 6