(2 . x + 1)^ 3 = 125
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a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
![](https://rs.olm.vn/images/avt/0.png?1311)
1. \(\left(x+1\right)^3-125\)
\(=\left(x+1\right)^3-5^3\)
\(=\left(x+1-5\right).\left[\left(x+1\right)^2+\left(x+1\right).5+5^2\right]\)
2. \(\left(x+4\right)^3-64\)
\(=\left(x+4\right)^3-4^3\)
\(=\left(x+4-4\right).\left[\left(x+4\right)^2+\left(x+4\right).4+4^2\right]\)
3. \(x^3-\left(y-1\right)^3\)
\(=(x^3-y+1).\left[\left(x^2\right)+x.\left(y+1\right)+\left(y+1\right)^2\right]\)
\(\)4. \(\left(a+b\right)^3-c^3\)
\(=\left[\left(a+b\right)-c\right].\left[\left(a+b\right)^2+\left(a+b\right).c+c^2\right]\)
5. \(125-\left(x+2\right)^3\)
\(=5^3-\left(x+2\right)^3\)
\(=\left(5-x-2\right).\left[5^2+5.\left(x+2\right)+\left(x+2\right)^2\right]\)
6. \(\left(x+1\right)^3+\left(x-2\right)^3\)
\(=\left[\left(x+1\right)+\left(x-2\right)\right].\left[\left(x+1\right)^2-\left(x+1\right).\left(x-2\right)+\left(x-2\right)^2\right]\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2^x=32=>x=5\)
\(2^{x+1}=32=>x+1=5=>x=4\)
\(x^3=125=>x=5\)
\(\left(x+1\right)^3=125=>x+1=5=>x=4\)
3. x3 = 125 => x = 5
4. (x + 1)3 = 125
(x + 1)3 = 53
=> x + 1 = 5
=> x = 4
t i c k nhé!! 45457474678789886973463125135153152454654657657868768
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow4^{x-5}\cdot17=68\)
=>4^x-5=4
=>x-5=1
=>x=6
b: \(\Leftrightarrow\dfrac{1}{3}:\left|2x-1\right|=\dfrac{1}{3}+\dfrac{2}{3}=1\)
=>|2x-1|=1/3
=>2x-1=1/3 hoặc 2x-1=-1/3
=>x=2/3 hoặc x=1/3
c: =>|2x-2|=|3x+15|
=>3x+15=2x-2 hoặc 3x+15=-2x+2
=>x=-17 hoặc x=-13/5
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(5\cdot x^3-5=0\)
`=> 5*x^3 = 0+5`
`=> 5*x^3 = 5`
`=> x^3 = 5 \div 5`
`=> x^3 = 1`
`=> x^3 = 1^3`
`=> x=1`
Vậy, `x=1.`
`b)`
\(( x+1)^2 = 16\)
`=> (x+1)^2 = (+-4)^2`
`=>`\(\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=4-1\\x=-4-1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy, `x \in {3; -5}`
`c)`
\(( x+1)^3 = 27\)
`=> (x+1)^3 = 3^3`
`=> x+1=3`
`=> x=3-1`
`=> x=2`
Vậy, `x=2.`
`d)`
\(( x-1)^3 = 343\)
`=> (x-1)^3 = 7^3`
`=> x-1=7`
`=> x=7+1`
`=> x=8`
Vậy, `x=8.`
`e)`
\((2x - 1^3) = 125\) hay đề là `(2x-1)^3 = 125` vậy ạ?
Mình làm cả 2 TH nhé!
`(2x-1^3)=125`
`=> 2x-1=125`
`=> 2x=125+1`
`=> 2x=126`
`=> x=126 \div 2`
`=> x=63`
TH2:
`(2x-1)^3 = 125`
`=> (2x-1)^3 = 5^3`
`=> 2x-1=5`
`=> 2x=5+1`
`=> 2x=6`
`=> x=6 \div 2`
`=> x=3`
Vậy, `x=3.`
(a) \(5x^3-5=0\Leftrightarrow5x^3=5\Leftrightarrow x^3=1\Leftrightarrow x=1\)
(b) \(\left(x+1\right)^2=16\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
(c) \(\left(x+1\right)^3=27\Leftrightarrow x+1=3\Leftrightarrow x=2\)
(d) \(\left(x-1\right)^3=343\Leftrightarrow x-1=7\Leftrightarrow x=8\)
(e) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)
=>(2x+1)^3=5^3
=>2x+1=5
=>2x=4
=>x=2
thx