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a) Ta có: ( 2 x + 1 ) 3 = 3 3 nên 2x + 1 = 3. Do đó x = 1.
b) Ta có: ( 2 x - 1 ) 3 = 5 3 nên 2x - 1 = 5. Do đó x = 3.
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(2x+1)^3=5^3
=>th1: 2x+1=5 =>. 2x=4 =>x=2
th2: 2x+1=-5 => 2x=-6 =>x=-3
vậy x=2 hoặc x=-3
( 2x + 1 )3 = 125
( 2x + 1 )3 = 53
=> 2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
Vậy x = 2
=))
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(2x+1)3=125
=> 2x + 1 = 41,666667
=> 2x = 42,66667
=> x = 21,33333
\(\left(2x+1\right)3=125\)
\(2x+1=125.3\)
\(2x+1=375\)
\(2x=375-1\)
\(2x=374\)
\(x=374:2\)
\(x=187\)
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`@` `\text {Ans}`
`\downarrow`
`2^x * 4 = 128`
`=> 2^x = 128 * 4`
`=> 2^x = 512`
`=> 2^x = 2^9`
`=> x = 9`
Vậy, `x = 9`
`x^15 = x`
`=> x^15 - x = 0`
`=> x(x^14 - 1) = 0`
`=>` TH1: `x = 0`
`TH2: x^14 - 1 = 0`
`=> x^14 = 1`
`=> x = 1`
Vậy, `x \in {0; 1}`
`(2x+1)^3 = 125`
`=> (2x+1)^3 = 5^3`
`=> 2x + 1 = 5`
`=> 2x = 5 - 1`
`=> 2x =4`
`=> x = 4 \div 2`
`=> x = 2`
Vậy,` x = 2.`
`(x - 5)^4 = (x-5)^6`
`=> (x-5)^4 - (x-5)^6 = 0`
`=> (x-5)^4 * [ 1 - (x-5)^2] = 0`
`=> - (x-6)(x-5)^4(x-4) = 0`
`TH1: (x - 5)^4 = 0`
`=> x - 5 = 0`
`=> x = 0 +5`
`=> x = 5`
`TH2: x - 6=0`
`=> x=6`
`TH3: x-4=0`
`=> x = 4`
Vậy, `x \in {4; 5; 6}`
a: =>2^x=32
=>x=5
b: =>x^15-x=0
=>x(x^14-1)=0
=>x=0; x=1;x=-1
c: =>2x+1=5
=>2x=4
=>x=2
d: =>(x-5)^4[(x-5)^2-1]=0
=>(x-5)(x-4)(x-6)=0
=>x=5;x=4;x=6
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
x6 : x3 = 125
x6-3 = 125
x3 = 125
=> x.x.x=125
Mà 5.5.5=125
=> x=5
Vậy x = 5
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow2x+1=15\)
\(\Leftrightarrow2x=14\Leftrightarrow x=7\)
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (2x +1)^3 =125
125 = 5^3
( 2.2 + 1 )^3 = 125
( 4 + 1 )^3 = 125
5^3 = 125
nên x = 2
nha bạn
(x - 1)^2 = (x - 1)^4
mình ko biết giải thích thế nào
x=0, x=1, x=2
nha bạn
![](https://rs.olm.vn/images/avt/0.png?1311)
b) 2 x - 1 3 = 125
2 x - 1 3 = 5 3
2x – 1 = 5
2x = 5 + 1
2x = 6
x = 6 : 2 = 3
(2x+1)3 = 125
<=> (2x+1)3 = 53
<=> 2x+1 = 5
<=> 2x = 4
<=> x = 2
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)