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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 \(\left[{}\begin{matrix}x=5\\x=4\\x=6\end{matrix}\right.\)
Bài 1:
a: =>13x+8=9x+20
=>4x=12
hay x=3
b: \(\Leftrightarrow5x-7=-8-11-3x\)
=>5x-7=-3x-19
=>8x=-12
hay x=-3/2
c: \(\Leftrightarrow\left[{}\begin{matrix}12x-7=5\\12x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{6}\end{matrix}\right.\)
e: =>3x+1=-5
=>3x=-6
hay x=-2
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