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a)\(\left(2x+1\right)^3=5^3\)
\(\Rightarrow2x+1=5\)
\(2x=5-1\)
\(2x=4\)
\(x=4:2\)
\(x=2\)
\(b,\left(2x-1\right)^3=125\)
\(\Rightarrow2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6:2=3\)
b) \(\left(2x-1\right)^3=125\)
\(\left(2x-1\right)^3=5^3\)
\(2x-1=5\)
\(2x=5+1\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
\(26\cdot5^{x-1}-5^{x-1}=125\)
\(\Rightarrow5^{^{x-1}}\cdot\left(26-1\right)=125\)
\(\Rightarrow5^{x-1}\cdot25=125\)
\(\Rightarrow5^{x-1}=\dfrac{125}{25}\)
\(\Rightarrow5^{x-1}=5\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=1+1\)
\(\Rightarrow x=2\)
\(\Leftrightarrow5^{x-1}\cdot\left(26-1\right)=125\)
=>\(5^{x-1}=5\)
=>x-1=1
=>x=2
`@` `\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) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
\(\left(2x+1\right)^3=125\\ \Rightarrow\left(2x+1\right)^3=5^3\\ \Rightarrow2x+1=5\\ \Rightarrow2x=4\\ \Rightarrow x=2.\\ b,\left(2x-1\right)^4=16\\ \Rightarrow\left(2x-1\right)^4=2^4\\ \Rightarrow2x-1=2\\ \Rightarrow2x=3\\ \Rightarrow x=\dfrac{3}{2}.\\ c,6.3^x-2.3^x=36\\ \Rightarrow3^x.\left(6-2\right)=36\\ \Rightarrow3^x.4=36\\ \Rightarrow3^x=9\\ \Rightarrow3^x=3^2\\ \Rightarrow x=2.\\ d,2^{x+1}-2^x=32\\ \Rightarrow2^x.\left(2-1\right)=32\\ \Rightarrow2^x=2^5\\ \Rightarrow x=5.\)
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
a,
\(\left(x-6\right)^2=9\\ \Rightarrow\left[{}\begin{matrix}x-6=-3\\x-6=3\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=9\end{matrix}\right.\)
b,
\(\left|x\right|=3\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
c,
\(\left|x+5\right|=15\\ \Rightarrow\left[{}\begin{matrix}x+5=-15\\x+5=15\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-20\\x=10\end{matrix}\right.\)
d,
\(2^{x-1}=16\\ \Rightarrow2^{x-1}=2^4\\ \Rightarrow x-1=4\\ \Rightarrow x=5\)
e,
\(5^{x+1}=125\\ \Rightarrow5^{x+1}=5^3\\ \Rightarrow x+1=3\\ \Rightarrow x=2\)
a: Ta có: \(\left(x-6\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=3\\x-6=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=3\end{matrix}\right.\)
b: Ta có: \(\left|x\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
c: Ta có: \(\left|x+5\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-20\end{matrix}\right.\)
K viết đề nha!
\(\left(x-1\right)^3\)= 125
=> x - 1 = 5
=> x = 6
Hội con 🐄 chúc bạn học tốt!!!
(x - 1)(x - 1)(x - 1) = 125
(x - 1)3 = 125
(x - 1)3 = 53
x - 1 = 5
x = 5 + 1
x = 6