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\(3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=3\left(1+3+3^2+3^3\right)+...+3^{2009}\left(1+3+3^2+3^3\right)\)
\(=40\left(3+...+3^{2009}\right)⋮40\)
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a. ( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = 72
<=> 2x + 1 = 7
<=> x = 3
b. ( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = 34
<=> 2x - 1 = 3
<=> x = 2
c. ( x + 1 )3 = 2x3
<=> x + 1 = 2x
<=> x = 1
d. ( 2x + 1 )3 = 3x3
<=> 2x + 1 = 3x
<=> x = 1
( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = ( ±7 )2
<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = ( ±3 )4
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
( x + 1 )3 = ( 2x )3
<=> x + 1 = 2x
<=> x - 2x = -1
<=> -x = -1
<=> x = 1
( 2x + 1 )3 = ( 3x )3
<=> 2x + 1 = 3x
<=> 2x - 3x = -1
<=> -x = -1
<=> x = 1
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2x x 16 = 128
2x = 128 : 16
2 x = 8
2x = 23
3x : 9 = 27
3x = 27 x 9
3x =243
3x = 35
[ 2x + 1 ]3 = 27
2x3 + 13 = 27
2x3 +1 = 27
2x3 = 27 - 1
2x3 = 26
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x(x + 3) = 0
=> \(\orbr{\begin{cases}x=0\\x+3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=0-3\end{cases}}\)
=> \(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)
(x - 2) (5 - x) = 0
=> \(\orbr{\begin{cases}x-2=0\\5-x=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0+2\\x=5-0\end{cases}}\)
=> \(\orbr{\begin{cases}x=2\\x=5\end{cases}}\)
(x - 1) (x2 + 1) = 0
=> \(\orbr{\begin{cases}x-1=0\\x^2+1=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=0+1\\x^2=0-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x^2=-1\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
x(x+3) = 0
→ x = 0
hoặc x + 3 = 0
⇒ x = 0
hoặc x = -3
Vậy x ∈ { 0 ; -3 }
( x -2 ) ( 5 -x ) = 0
⇒ x - 2 = 0
hoặc 5 - x = 0
⇒ x = 2
hoặc x= 5
Vậy x∈ { 2 ; 5 }
![](https://rs.olm.vn/images/avt/0.png?1311)
2.x2 = 18
x2 = 18 : 2
x2 = 9
x2 = 32
\(\Rightarrow\)x = 3
( 20 + 21 + 22 ) . x - 3 = 18
=> ( 1 + 2 + 4 ) . x = 21
=> 7 . x = 21
=> x = 21 : 7 = 3
2 . x2 = 18
=> x2 = 9
=> x2 = 32 = ( -3 )2
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(x^3-4x=0\)
\(\Leftrightarrow x\left(x^2-4\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow x=0hoacx-2=0hoacx+2=0\)
<=>x=0 hoặc x=2 hoặc x=-2
Vậy...
\(x^3-4x=0\)
\(x\left(x^2-4\right)=0\)
\(TH1:x=0\)
\(TH2:x^2-4=0\Rightarrow x^2=4\Rightarrow x=2,x=-2\)