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x=2;-4
x=5
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x=3;-4
x=5
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\(2^{x+2}-2^x=96\)
\(\Rightarrow2^x\cdot2^2-2^x=96\)
\(\Rightarrow2^x\left(2^2-1\right)=96\)
\(\Rightarrow2^x\left(4-1\right)=96\)
\(\Rightarrow2^x\cdot3=96\)
\(\Rightarrow2^x=96:3\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
\(5^x+5^{x+1}=750\)
\(\Rightarrow5^x+5^x\cdot5=750\)
\(\Rightarrow5^x\left(1+5\right)=750\)
\(\Rightarrow5^x\cdot6=750\)
\(\Rightarrow5^x=750:6\)
\(\Rightarrow5^x=125\)
\(\Rightarrow5^x=5^3\)
\(\Rightarrow x=3\)
\(2^{x+3}+2^x=144\)
\(\Rightarrow2^x\cdot2^3+2^x=144\)
\(\Rightarrow2^x\left(2^3+1\right)=144\)
\(\Rightarrow2^x\cdot9=144\)
\(\Rightarrow2^x=144:9\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
a/2x+2-2x=96
2x(22-1)=96
2x3=96
2x=96:3
2x=32
2x=25
x=5
b/(x-1)3=125
(x-1)3=53
x-1=5
x=6
c/(2x+1)3=343
(2x+1)3=73
2x+1=7
2x=6
x=3
\(3^{x+1}=9^x\)
\(\Leftrightarrow3^{x+1}=3^{2x}\)
\(\Leftrightarrow x+1=2x\)
\(\Leftrightarrow x+1=x+x\)
\(\Leftrightarrow x=1\)
Vậy \(x=1\)
\(2^{3x+2}=4^{x+5}\)
\(\Leftrightarrow2^{3x+2}=2^{2x+10}\)
\(\Leftrightarrow3x+2=2x+10\)
\(\Leftrightarrow3x=2x+8\)
\(\Leftrightarrow x=8\)
Vậy \(x=8\)
\(2^{x+2}-2^x=96\)
\(\Leftrightarrow2^x.2^2-2^x.1=96\)
\(\Leftrightarrow2^x\left(2^2-1\right)=96\)
\(\Leftrightarrow2^x.3=96\)
\(\Leftrightarrow2^x=\frac{96}{3}\)
\(\Leftrightarrow2^x=32\)
\(\Leftrightarrow2^x=2^5\)
\(\Leftrightarrow x=5\)
Vậy \(x=5\)
a) (9 - x)^3 = 216
(9 - x)^3 = 6^3
9 - x = 6
-x = 6 - 9
-x = -3
x = 3
b) 2^x + 2 - 2^x = 96
3.2^x = 96
2^x = 96 : 3
2^x = 32
2^x = 2^5
x = 5
a) \(2^x\cdot2^2-2^x=96\)
\(2^x\cdot\left(4-1\right)=96\)
\(2^x=32=2^5\)
=> x = 5
b) \(\left(2x+1\right)^3=7^3\)
=> 2x +1 = 7
=> 2x = 6
=> x = 3
Vậy,..........
a. 2x+2 - 2x = 96
2x . 22 - 2x = 96
2x. (22 - 1) = 96
2x = 96 : 3
2x = 32
2x = 25
x = 5
b. (2x + 1)3 = 343
(2x + 1)3 = 73
2x + 1 = 7
2x = 7 - 1
x = 6 : 2
x = 3
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) \(3^{4x+4}=81\Leftrightarrow3^4\cdot3^x=3^4\Leftrightarrow3^x=1\Leftrightarrow x=0\)
\(a,2^x+2^{x+1}=96\)
\(\Rightarrow2^x+2^x.2=96\) \(\Rightarrow2^x\left(1+2\right)=96\)
\(\Rightarrow2^x.3=96\) \(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\Rightarrow x=5\)
\(b,3^{4x+4}=81^{x+3}\)
\(\Rightarrow3^{4x+4}=3^{4x+12}\)
\(\Rightarrow4x+4=4x+12\) (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/ \(3^{4x+4}=81^{x+3}\)
\(\Rightarrow3^{4x+4}-81^{x+3}=0\)
\(3^{4x}.3^4-3^{4x}\cdot81^3=0\)
\(3^{4x}\cdot\left(81-81^3\right)=0\)
\(3^{4x}=\frac{0}{81-81^3}\)
\(3^{4x}=0\Rightarrow x=0\)
2x + 2x + 1 = 96
2x + 2x . 2 = 96
2x . (1 + 2) = 96
2x . 3 = 96
2x = 96 : 3
2x = 32
2x = 25
x = 5