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tìm x E Z biết
a, 0 : x =0
\(\Rightarrow x=\frac{0}{0}\)
\(\Rightarrow x\in\varnothing\)
b, 4 mũ x =64
\(\Rightarrow4^x=4^3\)
\(\Rightarrow x=3\)
c, 2 mũ x =16
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
d, 9 mũ x-1=9
\(\Rightarrow x-1=1\)
\(\Rightarrow x=2\)
e,x mũ 4 =16
\(\Rightarrow x^4=2^4\)
\(\Rightarrow x=2\)
g, 2 mũ x : 2 mũ 5 =1
\(\Rightarrow2^{x-5}=1\)
\(\Rightarrow x-5=0\)
\(\Rightarrow x=5\)
giúp mk với mk đang cần
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Ta có: \(4^{2x+4}=4^3\)
\(\Rightarrow\)\(2x+4=3\)
\(\Rightarrow\)\(x=\frac{-1}{2}\)
Ta có: \(3^{x-1}+3^{x-2}=108\)
\(\Rightarrow\)\(3^{x-1}+3^{x-2}=2^2.3^3\)
\(\Rightarrow\)\(3^{x-2}=3^3\)
\(\Rightarrow\)\(x-2=3\)
\(\Rightarrow x=5\)
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a)2^x + 1 . 2^2009 = 2^2010
=> 2^x + 1 + 2009 = 2^2010
=>2^x + 2010 = 2^2010
=>x + 2010 = 2010
=>x = 2010 - 2010 = 0
b)Chắc ý bạn là 6^17 : 6^15 + 44 : 11 đúng không?
Nếu thế thì mình sẽ giải như sau:
7x - 2x = 6^17 : 6^15 + 44 : 11
7x - 2x = 6^17 - 15 + 44 : 11
7x - 2x = 6^2 + 44 : 11
7x - 2x = 6^2 + 4
7x - 2x = 36 + 4
7x - 2x = 40
(7 - 2)x = 40
5x = 40
x = 40 : 5
x = 8
c)0 : x = 0
=>x ϵ N*
3^x = 9
3^x = 3^2
=> x = 2
d) x^4 = 16; 2^x : 2^5 = 1
x^4 = 2^4
x = 2
2^x : 2^5 = 1
2^x : 2^5 = 2^0
2^x - 5 = 2^0
=>x - 5 = 0
=>x = 0 + 5 = 5
e)|x - 2|= 0
<=> x - 2 = 0
<=> x = 0 + 2
<=> x = 2
g)4^x = 64
4^x = 4^3
x = 3
9^x - 1 = 9
9^x - 1 = 9^1
x - 1 = 1
x = 1 + 1
x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
1a.4,9,16,25,36,49,64,91,110,121,144,169,400,900,1600
b. 8,27,64,125,1000
C1111^2=1234321
2. x=6 ,x=3,x=0, x k thuộc n, x k thuộc n
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(-5\right).\left|x\right|=-75\)
\(\left|x\right|=\frac{-75}{-5}=15\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy....
\(b,\left(-6\right)^3.x^2=-1944\)
\(-216.x^2=-1944\)
\(x^2=9\)
\(\Rightarrow x=\pm3\)
Vậy....
\(d,\left|9-x\right|=-7+64\)
\(\left|9-x\right|=57\)
\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)
Vậy...
\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)
\(\left|x+101\right|+16=215\)
\(\left|x+101\right|=199\)
\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)
Vậy..
hok tốt!!
a,\(\left(-5\right).\left|x\right|=-75\)
\(=>\left|x\right|=-75:\left(-5\right)=15\)
\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
b,\(\left(-6\right)^3.x^2=-1944\)
\(=>\frac{1944}{216}=x^2\)
\(=>x=\sqrt{\frac{1944}{216}}=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
5.3x=405
3x=405:5
3x=81
3x=34
Vậy x=4
2x:8=4
2x=4.8
2x=32
2x=25
Vậy x=5
x28=x5
x^28-x^5=0
x^5.x^23-x^5.1=0
x^5.(x^23-1)=0
suy ra x^5=0 hoặc x^23-1=0 suy ra x^5=0^5 hoặc x^23=0+1=1 suy ra x=0 hoặc x^23=1^23 suy ra x=0 hoăc x=1
9
(x-2)^4=256
(x-2)^4=4^4
x-2=4
x=4+2=6
(x+1)^3=125
(x+1)^3=5^3
x+1=5
x=5-1=4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^x=32\)=>\(2^x=2^5\)=>\(x=5\)
b) \(64\times4^x=4^5\)
\(4^3.4^x=4^5\)
\(4^{3+x}=4^5\)
=>\(3+x=5\)
=> x=2
c)\(2^x-15=17\)
\(2^x=17+15\)
\(2^x=32\)
=>\(2^x=2^5\)
=>\(x=5\)
a, 2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
b, 64 x 4 mũ x = 4 mũ 5
64 x 4 mũ x = 1024
4 mũ x = 1024 : 64
4 mũ x = 16
4 mũ x = 4 mũ 2
x = 2
c, 2 mũ x - 15 = 17
2 mũ x = 17 + 15
2 mũ x = 32
2 mũ x = 2 mũ 5
x = 5
4x-1-64=0
4x-1=64=43
=>x-1=3
x=4
\(4^{x-1}-64=0\)
\(\Leftrightarrow4^{x-1}=64\)
\(\Leftrightarrow4^{x-1}=4^3\)
\(\Rightarrow x-1=3\Leftrightarrow x=4\)