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a) (2^x).4=128
2^x = 128:4
2^x = 32
mà 32=2^5=>x=5
b) ta có: x^15=x
theo quy ước: 0^15=0;1^15=1
=> x=1
4 câu còn lại mai mình sẽ giải nhé
2x.4=128
=> 2x=128:4
=> 2x=32
=> 2x=25
=> x=5
(2x+1)3=125
=>(2x+1)3=53
=>2x+1=5
=>2x=5-1
=>2x=4
=>2x=22
=>x=2
2x-15=17
=>2x=17+15
=>2x=32
=>2x=25
=>x=5
\(a,2^x\cdot4=128\)
\(2^x=128:4=32\)
\(2^x=2^5\)
\(x=5\)
\(b,x^{15}=x\)
\(x^{15}-x=0\)
\(x\left(x^{14}-1\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x^{14}-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(c,16^x< 128\)
\(2^{4x}< 2^7\)
\(4x< 7\)
\(x=1\)
d,\(5^x\cdot5^{x+1}\cdot5^{x+2}< 1000000000000000000:2^{18}\)
\(5^{x+x+1+x+2}< 10^{18}:2^{18}\)
\(5^{3x+3}< 5^{18}\)
\(3x+3< 18\)
\(3\left(x+1\right)< 18\)
\(x+1< 6\)
\(x< 5\)
\(e,2^x\cdot\left(2^2\right)^2=\left(2^3\right)^2\)
\(2^x\cdot2^4=2^6\)
\(2^{x+4}=2^6\)
\(x+4=6\)
\(x=2\)
\(f,\left(x^5\right)^{10}=x\)
\(x^{50}=x\)
\(x^{50}-x=0\)
\(x\left(x^{49}-1\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x^{49}-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(2^x.4=128\)
\(\Rightarrow2^x=128:4\)
\(\Rightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
tíc mình nha
Ta có: A = 1 + 2 + 22 + 23 + 24 + ...... + 2100
=> 2A = 2 + 22 + 23 + 24 + ...... + 2101
=> 2A - A = 2101 - 1
=> A = 2101 - 1
(2*x+1)3=125
(2*x+1)=5^3
2*x+1 =5
2*x =5-1
2*x =4
x=4:2
x=2
con may cau hoi kia lat lam tiep
chuc bn hoc gioi!
nha
a) \(2^x\times4=128\)
\(2^x=128:4=32=2^5\)
\(x=5\)
b) \(x^{100}=x\)
\(x^{100}-x=0\)
\(x\left(x^{99}-1\right)=0\)
x=0 hoặc x=1
c) \(\left(2x+1\right)^3=125=5^3\)
\(2x+1=5\)
\(x=2\)
d) \(\left(x-2\right)^{2016}=\left(x-2\right)^{2014}\)
\(\left(x-2\right)^{2014}\left(\left(x-2\right)^2-1\right)=0\)
\(x=0\) hoặc \(\left(x-2\right)^2=1\)
x=0 hoặc x=3 hoặc x=1
a)2x.4=128
2x=128:4=32
=>x=5
b)x100=x
=>x=1
c) (2x+1)3 =125
(2x+1)3=53
=> 2x+1=5
2x=5-1=4
x=4:2
x=2
d) (x-2)2016=(x-2)2014
=> x=2 (vì 2-2=1,mà 1 mũ mấy cũng bằng 1)
2^(2+x)=2^7
2+x=7
x=5
(x-1)^3-(x-1)^2=0
(x-1)^2.(x-1-1)=0
(x-10)^2.(x-2)=0
=>+)x-10=0=>x=10
+)x-2=0 =>x=2
\(2^x\cdot4=128\)
\(2^x=32=2^5\)
Vậy x = 5
\(\left(x-1\right)^2=\left(x-1\right)^3\)
\(\left(x-1\right)^2-\left(x-1\right)^3=0\)
\(\left(x-1\right)^2-\left(x-1\right)^2\cdot\left(x-1\right)=0\)
\(\left(x-1\right)^2\left(1-x+1\right)=0\)
\(\left(x-1\right)\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)