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a: =>1/3:x=3/5-2/3=9/15-10/15=-1/15
=>x=-1/3:1/15=5
b: \(\Leftrightarrow x\cdot\dfrac{2}{3}-3=\dfrac{2}{5}\cdot\left(-10\right)=-4\)
=>x*2/3=-1
=>x=-3/2
c: =>2x+1=4 hoặc 2x+1=-4
=>x=3/2 hoặc x=-5/2
h: =>x-3=4
=>x=7
g: =>2x-1=3
=>2x=4
=>x=2
f: \(\Leftrightarrow x\cdot\left(\dfrac{3}{2}-\dfrac{7}{3}\right)=\dfrac{3}{2}-\dfrac{2}{3}\)
=>x*-5/6=5/6
=>x=-1
d: =>|2x-1|=3
=>2x-1=3 hoặc 2x-1=-3
=>x=-1 hoặc x=2
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32x+1=81 : 3=27
32x+1=33
2x+1=3
2x=3-1
2x=2
x=2:2
x=1
vay x=1
k nha
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a, => 3^2x = 3^4
=>2x=4
=>x=2
b,=>2x+3x-3=17
=>5x-3=17
=>5x=17+3=20
=>x=4
c,=>x+15=(2x+3).2 = 4x+6
=>4x+6-(x+15)=0
=>4x+6-x-15=0
=>3x-9=0
=>3x=9
=>x=3
k mk nha
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\(4.\left(3-2x\right)^2=\frac{1}{81}\)
\(\left(3-2x\right)^2=\frac{1}{324}\)
\(\Leftrightarrow\orbr{\begin{cases}3-2x=\frac{1}{18}\\3-2x=\frac{-1}{18}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{53}{36}\\x=\frac{55}{36}\end{cases}}}\)
Vậy \(x\in\left\{\frac{53}{36};\frac{55}{36}\right\}\)
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a, 2x-1 = 64
=> 2x-1 = 26
=> x - 1 = 6
=> x = 6 + 1
=> x = 7
b, 32x-10 = 81
=> 32x-10 = 34
=> 2x - 10 = 4
=> 2x = 14
=> x = 7
c, 22x+3 = 1024
=> 22x+3 = 210
=> 2x + 3 = 10
=> 2x = 7
=> x = \(\frac{7}{2}\)
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3x . 812x + 1 = 81
3x . 38x + 4 = 81
3x + 8x + 4 = 34
=> x + 8x + 4 = 4
=> x + 8x = 0
=> 9x = 0
=> x = 0
Vậy x = 0
\(3^x.81^{2x+1}=81\)
\(\Rightarrow3^x.3^{8x+4}=81\)
\(\Rightarrow3^{9x+4}=3^4\)
\(\Rightarrow9x+4=4\Leftrightarrow x=0\)
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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
\(\Leftrightarrow2x+1=3\)
hay x=1
\(\left(2x+1\right)^3=27\)
\(\left(2x+1\right)=3\)
\(2x=2\)
\(x=1\)
Vậy \(x=1\)