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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.
(2x+1)3=125
=> 2x + 1 = 41,666667
=> 2x = 42,66667
=> x = 21,33333
\(\left(2x+1\right)3=125\)
\(2x+1=125.3\)
\(2x+1=375\)
\(2x=375-1\)
\(2x=374\)
\(x=374:2\)
\(x=187\)
`@` `\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
\(\left(2x+1\right)^3=125\)
\(\Leftrightarrow2x+1=15\)
\(\Leftrightarrow2x=14\Leftrightarrow x=7\)
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(x=2\)
b) 2 x - 1 3 = 125
2 x - 1 3 = 5 3
2x – 1 = 5
2x = 5 + 1
2x = 6
x = 6 : 2 = 3
a) \(\left(2x-1\right)^3=27\)
\(\Rightarrow\left(2x-1\right)^3=3^3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=3+1=4\)
\(\Rightarrow x=4:2=2\)
b) \(\left(2x+1\right)^3=125\)
\(\Rightarrow\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\)
\(\Rightarrow x=4:2=2\)
c) \(\left(x+2\right)^3=\left(2x\right)^3\)
\(\Rightarrow x+2=2x\)
\(\Rightarrow2=2x-2=1x\)
\(\Rightarrow x=2:1=2\)
d) \(\left(2x-1\right)^7=x^7\)
\(\Rightarrow2x-1=x\)
\(\Rightarrow-1=x-2x\)
\(\Rightarrow-1=-x\)
\(\Rightarrow x=1\)
x=2
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