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\(26\cdot5^{x-1}-5^{x-1}=125\)
\(\Rightarrow5^{^{x-1}}\cdot\left(26-1\right)=125\)
\(\Rightarrow5^{x-1}\cdot25=125\)
\(\Rightarrow5^{x-1}=\dfrac{125}{25}\)
\(\Rightarrow5^{x-1}=5\)
\(\Rightarrow x-1=1\)
\(\Rightarrow x=1+1\)
\(\Rightarrow x=2\)
\(\Leftrightarrow5^{x-1}\cdot\left(26-1\right)=125\)
=>\(5^{x-1}=5\)
=>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) \(\left(2x-1\right)^3=125\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\)
3x=234—> 243 chứ bạn
3x=243
Vì 35=243
Nên x=5
3x=4096
6x=216
Vì 63=216
Nên x=3
5.3x=405
3x=405:5
3x=81
Vì 34=81
Nên x=4
2x=4
Vì 22=4
Nên x=2
x5=32
Vì 25=32
Nên x=5
x4=81
Vì 34=81 và (-3)4=81
Nên x=3 hoặc x=-3
(x—2)5=243
(x—2)5=35
Nên x—2=3
x=3+2
x=5
(x—1)4=16
(x—1)4=24
Nên x—1 =2 hoặc x—1=-2
x=2+1 hoặc x=-2 +1
x=3 hoặc x=-1
(x+1)3=125
(x+1)3=53
Nên x+1=3
x=3-1
x=1
(x—2)3=64
(x—2)3=43
Nên x—2=4
x=4+2
x=6
(x—1)5=32
(x—1)5=25
==> x—1=5
x=5+1
x=6
\(3^x=243\Rightarrow3^x=3^5\Rightarrow x=5\)
câu thứ 2 mn chịu
\(6^x=216\Rightarrow6^x=6^3\Rightarrow x=3\)
\(5.3^x=405\\ 3^x=405:5\\ 3^x=81\\ \Rightarrow3^x=3^4\\ \Rightarrow x=4\)
\(2^x=4\\ \Rightarrow2^x=2^2\\ \Rightarrow x=2\)
\(x^5=32\\ \Rightarrow x^5=2^5\\ \Rightarrow x=2\)
câu 7 mn ko hỉu
\(x^4=81\\ x^4=\left(+-3\right)^4\\ x=+-3\)
câu 9 mn chịu
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: \(\left|x+5\right|=15\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-20\end{matrix}\right.\)
\(a,4^{2x-6}=1\)\(\Leftrightarrow2x-6=0\Leftrightarrow2x=6\Rightarrow x=3\)
\(b,2^{2x-1}=16\Rightarrow2^{2x-1}=2^4\)
\(\Rightarrow2x-1=4\Rightarrow2x=5\Rightarrow x=\frac{5}{2}\)
\(c,5< 5^x< 125\Rightarrow5^1< 5^x< 5^3\)\(\Rightarrow1< x< 3\)
\(d,5^{x+1}=125\Rightarrow5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2\)
\(a,4^{2x-6}=1\)
\(4^{2x-6}=4^0\)
\(\Rightarrow2x-6=0\)
\(\Rightarrow2x=6\Leftrightarrow x=3\)
\(b,2^{x-1}=16\)
\(2^{x-1}=2^4\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
a. 3^x=1-x^2
x=0 la nghiem
x>=1; VT>=3 VP<=0 vo nghiem
b. (de bai thieu n khac 0 vi neu n=0 dung voi moi x)
3x-14=1=> x=5
c.(5^2x5^x+1)=5^4
5^x+1=5^2=> x=1