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\(2^{x+3}.4^2=64\Leftrightarrow2^{x+3}.2^4=64\Leftrightarrow2^{x+7}=2^6\Leftrightarrow x+7=6\Leftrightarrow x=-1\)
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(428-4x) .27=1728
<=> 428-4x = 1728 : 27
<=> 428-4x = 64
<=> 4x = 428 - 64
<=> 4x = 364
<=> x= 364 :4
<=> x= 91
Vậy x= 91
x.x2.x3=64
<=> x1+2+3 = 64
<=> x6 = 64
<=> x6= (+ - 2 )6
=> x = cộng trừ 2
Vậy x= cộng trừ 2
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a,\(\left(3x-4\right).2^3=64\)
(=) \(\left(3x-4\right)=64:8\)
(=) \(3x-4=8\)
(=) \(3x=8+4\)
(=) \(3x=12\)
(=) \(x=12:3\)
(=) \(x=4\)
b, Ta có \(2^3.3^2-4x=4^3\)
(=) \(8.9-4x=64\)
(=) \(72-4x=64\)
(=) \(4x=72-64\)
(=) \(4x=8\)
(=) \(x=8:4\)
(=) \(x=2\)
a) ( 3x - 4 ) . 23 = 64
=> ( 3x - 4 ) = 64 : 23
=> ( 3x - 4 ) = 26 : 23
=> ( 3x - 4 ) = 23
=> 3x - 4 = 8
=> 3x = 12
=> x = 4
b) 23 . 32 - 4x = 43
=> 8 . 9 - 4x = 64
=> 72 - 4x = 64
=> 4x = 8
=> x = 2
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a) (x+3).213.(x+3)=217
(x+3)2=217:213
(x+3)2=24
(x+3)2=(22)2
x+3=22
x+3=4
x=1
b) (43+64).(3x+4x+5x)=3072
128.12x=3072
12x=3072:128
12x=24
x=2
c) (x-2)3=(x-2)2
=> x-2 = 0
x=2
hoặc x-2=1
x=3
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a) (x-3)100-1=0
\(\Leftrightarrow\left(x-3\right)^{100}=0+1=1\)
\(\Leftrightarrow\left(x-3\right)^{100}=1^{100}=\left(-1\right)^{100}\)
\(\Rightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1+3=4\\x=\left(-1\right)+3=2\end{matrix}\right.\)
Vậy: x\(\in\left\{4;2\right\}\)
b) (9-5x)2019+1=0
\(\Leftrightarrow\left(9-5x\right)^{2019}=0-1=-1=\left(-1\right)^{2019}\)
\(\Rightarrow9-5x=-1\)
\(\Leftrightarrow5x=9-\left(-1\right)=10\)
\(\Leftrightarrow x=10:5=2\)
Vậy: x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^2.2^4.x=16^2\)
\(2^6x=\left(2^4\right)^2\)
\(2^6.x=2^8\)
\(x=2^8:2^6\)
\(x=2^2\)
\(x=4\)
vay \(x=4\)
b) \(5^{2x+1}=125\)
\(5^{2x+1}=5^3\)
\(\Rightarrow2x+1=3\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
vay \(x=1\)
c) \(\left(2x-1\right)^3=64\)
\(\left(2x-1\right)^3=4^3\)
\(\Rightarrow2x-1=4\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=\frac{5}{2}\)
vay \(x=\frac{5}{2}\)
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`(4x-2)^6 = 64`
`(4x-2)^6 = ` \(\left(\pm2\right)^6\)
`@TH1:`
`4x-2=2`
`4x=2+2`
`4x=4`
`x=4:4`
`x=1`
`@TH2:1`
`4x-2=-2`
`4x=-2+2`
`4x=0`
`x=0:4`
`x=0`
Vậy `x={0;1}`
( 4x - 2 )6 = 64
( 4x - 2 )6 = 26
4x - 2 = 2
4x = 2 + 2
4x = 4
x = 1