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a) \(2^x=32\)
Ta có: \(2^5=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
b) Sửa đề tí: \(9< 3^x< 81\)
\(\Rightarrow3^2< 3^x< 3^4\)
\(\Rightarrow2< x< 4\)
\(\Rightarrow x=\left\{3\right\}\)
Vậy x = 3
c) Ta có: \(25\le5^x\le125\)
\(\Rightarrow5^2\le5^x\le5^3\)
\(\Rightarrow2\le x\le3\)
\(\Rightarrow x=\left\{2;3\right\}\)
Vậy x = 2 hoặc x = 3
d) \(\left(x-2\right)^3\times5=40\)
\(\Rightarrow\left(x-2\right)^3=8\)
Mà \(8=2^3\Rightarrow\left(x-2\right)^3=2^3\)
Suy ra: x - 2 = 2
Vậy x = 4
a, ( x-11 )5=15
=> x- 11 = 1
x = 12
b, ( 24 -x)3=23
24-x = 2
x = 22
c, ( 2x + 1 )3=53
2x + 1 = 5
2x = 4
x = 2
lik e mình nha minion
Bài 1 :
a/ \(a^3.a^9=a^{3+9}=a^{12}\)
b/\(\left(a^5\right)^7=a^{5.7}=a^{35}\)
c/ \(\left(a^6\right).4.a^{12}=a^{24}.a^{12}.4=a^{24+12}.4=a^{36}.4\)
d/ \(\left(2^3\right)^5.\left(2^3\right)^3=2^{15}.2^9=2^{15+9}=2^{24}\)
e/ \(5^6:5^3+3^3.3^2\)
\(=5^3+3^5=125+243=368\)
i/ \(4.5^2-2.3^2\)
\(=2^2.5^2-2.3^2\)
\(=2^2.25-2^2.14\)
\(=2^2.\left(25-14\right)\)
\(=2^2.11\)
\(=4.11=44\)
a)(x-1)^3=125=5^3
=>x-1=5=>x=6
b)2^x+2-2^x=96
=>2^x.2^2-2^x=96
=>2^x.(2^2-1)=96
=>2^x.3=96
=>2^x=96/3=32=2^5
=>x=5
c)(2^x +1)^3=343=7^3
=>2^x +1=7
=>2^x=6
sai đề ko?
Đặt \(A=5+5^3+5^5+....+5^{47}+5^{49}\)
\(\Rightarrow5^2A=5^3+5^5+5^7+.....+5^{49}+5^{51}\)
\(\Rightarrow5^2A-A=\left(5^3+5^5+5^7+....+5^{49}+5^{51}\right)-\left(3+3^3+3^5+....+5^{47}+5^{49}\right)\)
\(\Rightarrow24A=5^{51}-5\)
\(\Rightarrow A=\dfrac{5^{51}-5}{24}\)
Vậy ............................................................
1)a) \(\left(3x-7\right)^5=32\Rightarrow\left(3x-7\right)^5=2^5\)
\(\Rightarrow3x-7=2\Rightarrow3x=9\Rightarrow x=3\)
Vậy \(x=3\)
b) \(\left(4x-1\right)^3=-27.125\)
\(\Rightarrow\left(4x-1\right)^3=-3^3.5^3=-15^3\)
\(\Rightarrow4x-1=-15\Rightarrow4x=-14\Rightarrow x=-3,5\)
Vậy \(x=-3,5\)
c) \(3^{4x+4}=81^{x+3}\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\)
\(\Rightarrow4x=4x+8\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)^7=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^7-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^7.\left[1-\left(x-5\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^7=0\\1-\left(x-5\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\\left(x-5\right)^2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=-1\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
a: \(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5+3^5}\cdot\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5+2^5+2^5+2^5+2^5}=2^x\)
\(\Leftrightarrow2^x=\dfrac{4^5}{3^5}\cdot\dfrac{6^5}{2^5}=4^5=2^{10}\)
=>x=10
b: \(\left(x-1\right)^{x+4}=\left(x-1\right)^{x+2}\)
\(\Leftrightarrow\left(x-1\right)^{x+2}\left[\left(x-1\right)^2-1\right]=0\)
\(\Leftrightarrow x\left(x-1\right)^{x+2}\cdot\left(x-2\right)=0\)
hay \(x\in\left\{0;1;2\right\}\)
c: \(6\left(6-x\right)^{2003}=\left(6-x\right)^{2003}\)
\(\Leftrightarrow5\cdot\left(6-x\right)^{2003}=0\)
\(\Leftrightarrow6-x=0\)
hay x=6
a) 3x-5=4
3x = 4+5
3x=9
=> 9 = 3x = 32
<=> x=2
b) 5x-2 = 125
=> 5x-2 = 53
<=> x-2 = 3
x = 3+2
x = 5
c) x3 = 8
<=> x3 = 8 = 23
<=> x= 2
d) (x-1)3 = 125
=> x-1 = 5
x = 5+1
x = 6
a) 3^x-5=4
=>3^x =9
=>3^x =3^2
=> x = 2
b)5^x-2=125
=>5^x-2=5^3
=>x-2=3
=>x =5
a,3^x=4+5
3^x=9
3^x=3^2
=>x=2
b,5^x-2=125
5^x-2=5^3
=>x-2=3
x=3+2
x=5
c,x^3=8
x^3=2^3
=>x=2
d,(x-1)^3=125
(x-1)^3=5^3
=>x-1=5
x=5+1
x=6
\(a.\)
\(3^x-5=4\)
\(3^x=9\)
\(3^x=3^3\)
\(x=3\)
\(b.\)
\(5^{x-2}=125\)
\(5^{x-2}=5^3\)
\(x-2=3\)
\(x=5\)
\(c.\)
\(x^3=8\)
\(x^3=2^3\)
\(x=2\)
\(d.\)
\(\left(x-1\right)^3=125\)
\(\left(x-1\right)^3=5^3\)
\(x-1=5\)
\(x=6\)
a,\(3^x-5=4\)
\(3^x=4+5\)
\(3^x=9\)
\(3^x=3^2\)
\(\Rightarrow x=2\)
Vậy x=2
b,\(5^{x-2}=125\)
\(5^{x-2}=5^3\)
\(\Rightarrow x-2=3\)
x=2+3
x=5
Vậy x=5
c,\(x^3=8\)
\(x^3=2^3\)
\(\Rightarrow x=2\)
Vậy x=2
d,\(\left(x-1\right)^3=125\)
\(\left(x-1\right)^3=5^3\)
\(\Rightarrow x-1=5\)
x=5+1
x=6
Vậy x=6
a) 3x - 5 = 4 b) 5x-2 = 125 c) x3 = 8 d) ( x - 1 )3 = 125
=> 3x = 9 => 5x-2 = 53 => x3 = 23 => ( x - 1 )3 = 53
=> 3x = 32 => x - 2 = 3 => x = 2 => x - 1 = 5
=> x = 2 => x = 5 => x = 6
a. \(3^x-5=4\)
\(3^x=4+5\)
\(3^x=9=3^2\)
Vậy x = 2
b. \(5^{x-2}=125\)
\(5^{x-2}=5^3\)
\(5^x-5^2=5^3\)
\(5^x=5^3+5^2\)
\(5^x=5^5\)
Vậy x = 5
a) 3^x -5=4
=>3^x=4+5
=>3^x=9
mà 3^2=9
=>3^x=9
=>x=2
b) ta có : 125=5^3
mà 5^x-2=125
=>x-2=3
=>X=3+2
=>x=5
c)ta có 8=2^3
=>x=2
d)ta có :5^3=125
=>x-1=5
=>x=5+1
=>x=6
a. 3x - 5 = 4
3x = 4 + 5
3x = 9
3x = 32
x = 2
b. 5x - 2 = 125
5x . 52 = 125
5x . 52 = 53
5x = 53 : 52
5x = 51
x = 1
c. x3 = 8
x3 = 23
x = 2
d. ( x - 1 )3 = 125
( x - 1 )3 = 53
x - 1 = 5
x = 5 + 1
x = 6
hay đó