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54 + ( - 8 ) + x = 3 . x
46 + x = 3 . x
46 = 3 . x - x
46 = 2 .x
x = 23
b , 3x + 8 chia hết cho x - 1
Mà x - 1 chia hết cho x - 1
=> 3 ( x - 1 ) chia hết cho x - 1
=> 3x - 3 chia hết cho x - 1
=> 3x + 8 - 3x - 3 chia hết cho x - 1
=> 5 chia hết cho x - 1
=> x - 1 \(\in\) Ư ( 5 )
=> x - 1 \(\in\) { 1 , -1 , 5 , -5 }
=> x \(\in\) { 2 ; 0 ; 6 ; -4 }
a) Ta có : 2 x : 2 2 = 2 5 nên x = 7.
b) Ta có: 3 x : 3 2 = 3 5 nên x = 7.
c) Ta có : 4 4 : 4 x = 4 2 nên x = 2.
d) Ta có : 5 x : 5 2 = 5 2 nên x = 4,
e) Ta có: 5 x + 1 : 5 = 5 4 nên x = 4.
f) Ta có : 4 2 x - 1 : 4 = 4 2 nên x = 2
$\Rightarrow 3^x(1+3+3^2+3^3)=1080$
$\Rightarrow 3^x.40=1080$
$\Rightarrow 3^x=27=3^3$
$\Rightarrow x=3$
1.
7.5.(-2)=12.(-2)=-24
-54-18=-72
2.
x+4=-12
x=-12-4
x=-16
Vậy x=-16
3.
3x-5=-7-13
3x-5=-20
3x=-20+5
3x=-15
x=-15:3
x=-5
Vậy x=-5
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
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: \(x=\dfrac{68}{189}\)
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: \(x=\dfrac{68}{189}\)
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: \(x=\dfrac{68}{189}\)
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: \(x=\dfrac{68}{189}\)
\(3^x-3^{x-1}=54\)
=>\(3^x-\dfrac{1}{3}\cdot3^x=54\)
=>\(\dfrac{2}{3}\cdot3^x=54\)
=>\(3^x=54:\dfrac{2}{3}=81=3^4\)
=>x=4