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\(a,2^x+2^{x+1}=96\)
\(\Rightarrow2^x+2^x.2=96\) \(\Rightarrow2^x\left(1+2\right)=96\)
\(\Rightarrow2^x.3=96\) \(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(b,3^{4x+4}=81^{x+3}\)
\(\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\) (Vô lý)
Vậy \(x\in\varnothing\)
a/ \(2^x+2^{x+1}=96\)
\(2^x+2^x.2=96\)
\(2^x\cdot\left(2+1\right)=96\)
\(2^x=\frac{96}{3}=32\)
\(2^x=2^5\)
\(=>x=5\)
b/ \(3^{4x+4}=81^{x+3}\)
\(\Rightarrow3^{4x+4}-81^{x+3}=0\)
\(3^{4x}.3^4-3^{4x}\cdot81^3=0\)
\(3^{4x}\cdot\left(81-81^3\right)=0\)
\(3^{4x}=\frac{0}{81-81^3}\)
\(3^{4x}=0\Rightarrow x=0\)
Câu 1:
\(\dfrac{128}{\left(n-3\right)^3}=2\)
\(\Leftrightarrow\left(n-3\right)^3=64\)
=>n-3=4
hay n=7
Câu 2:
\(\Leftrightarrow x^3+12=57\cdot4=228\)
\(\Leftrightarrow x^3=216\)
hay x=6
\(a,3^n=3^4\)
\(\Rightarrow n=4\)
\(b,2008^n=2008^0\)
\(\Rightarrow n=0\)
a, 541 + ( 218 - x ) = 735
=> 218 - x = 735 - 541
=> 218 - x = 194
=> x = 218 - 194
=> x = 24
Vậy x = 24
b, 96 - 3.( x + 1 ) = 42
=> 3.( x + 1 ) = 96 - 42
=> 3.( x + 1 ) = 54
=> x + 1 = 54 : 3
=> x + 1 = 18
=> x = 18 - 1
=> x = 17
Vậy x = 17
c, ( x - 36 ) : 18 = 12
=> x - 36 = 12.18
=> x - 36 = 216
=> x = 216 + 36
=> x = 252
Vậy x = 252
d, 3x + 3x+1 + 3x+2 = 13.81
=> 3x.1 + 3x.3 + 3x.32= 13.81
=> 3x.( 1 + 3 + 32 ) = 13.81
=> 3x.( 1 + 3 + 9 ) = 13.81
=> 3x.13 = 13.81
=> 3x= 13.81:13
=> 3x= 81
=> 3x = 34
=> x = 4
Vậy x = 4
a) \(2^x+2^{x+1}=96\Leftrightarrow2^x+2\cdot2^x=96\Leftrightarrow2^x\cdot3=96\Leftrightarrow2^x=48\)
Không có x nguyên thỏa mãn.
b) \(3^{4x+4}=81\Leftrightarrow3^4\cdot3^x=3^4\Leftrightarrow3^x=1\Leftrightarrow x=0\)
a) 2x + 2x+1 = 96
=> 2x(1 + 2) = 96
=> 2x.3 = 96
=> 2x = 96 : 3
=> 2x = 32
=> 2x = 25
=. x = 5
\(A,\left(a^6\right)^4.a^{12}=a^{24}.a^{12}=a^{36}\)
\(B,5^6:5^3+3^3.3^2=5^3+3^5=125+243=368\)
Tìm X
\(A,\left(x-1\right)^3=125=5^3\)
\(x-1=5\)
\(\Rightarrow x=6\)
\(B,720:\left[41-\left(2x-5\right)\right]=2^3.5=40\)
\(\Leftrightarrow41-\left(2x-5\right)=\frac{720}{40}=18\)
\(\Leftrightarrow2x-5=23\)
\(\Leftrightarrow x=\frac{28}{2}=14\)