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\(\Leftrightarrow\dfrac{5}{2}\cdot2\cdot2^x-2^x=384\)
\(\Leftrightarrow4\cdot2^x=384\)
\(\Leftrightarrow2^x=96\)
hay \(x\in\varnothing\)
a) \(5.2^{x+1}.2^{-2}-2^x=384\Leftrightarrow2^x\left(5.2^{-2}.2-1\right)=384\)\(\Leftrightarrow2^x.1,5=384\Leftrightarrow2^x=384:1,5=256=2^8\)
\(\Rightarrow x=8\)
b) \(3^{x+2}.5^y=45^x\Leftrightarrow3^{x+2}.5^y=3^{2x}.5^x\Leftrightarrow\frac{3^{2x}}{3^{x+2}}=\frac{5^y}{5^x}\)\(\Leftrightarrow3^{2x-x+2}=5^{y-x}\Leftrightarrow3^{x+2}=5^{y-x}\)
\(\Rightarrow x+2=y-x=0\Rightarrow x=y=-2\)
a) \(5.2^{x+1}.2^{-2}-2^x=384\)
\(\Leftrightarrow2^x.2.\frac{5}{4}-2^x=384\)
\(\Leftrightarrow2^x.\left(\frac{5}{2}-1\right)=384\)
\(\Leftrightarrow2^x.\frac{3}{2}=384\)
\(\Leftrightarrow2^x=256\)
\(\Leftrightarrow2^x=2^8\)
\(\Leftrightarrow x=8\)
c) \(\left(x+1\right)^{x+1}=\left(x+1\right)^{x+3}\)
\(\Leftrightarrow\left(x+1\right)^{x+3}-\left(x+1\right)^{x+1}=0\)
\(\Leftrightarrow\left(x+1\right)^{x+1}\left[\left(x+1\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{x+1}=0\\\left(x+1\right)^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\\left(x+1\right)^2=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x\in\left\{0;-2\right\}\end{cases}}}\)
Vậy \(x\in\left\{0;-1;-2\right\}\)
a. 2x-1+ 5.2x-1:2=7/32
=> 2x+1.(1+5/2)=7/32
=>2x+1.7/2=7/32
=> 2x+1=1/16=1/24
=> x+1=-4=>x=-5
b)\(2^{x-1}+5\cdot2^{x-2}=\frac{7}{32}\)
\(2^x:2+5\cdot2^x:2^2=\frac{7}{32}\)
\(2^x:2+2^x:\frac{4}{5}=\frac{7}{32}\)
\(2^x\cdot\left(\frac{1}{2}+\frac{5}{4}\right)=\frac{7}{32}\)
\(2^x\cdot\frac{7}{4}=\frac{7}{32}\)
\(2^x=\frac{7}{32}:\frac{7}{4}=\frac{1}{8}\)
\(2^x=\frac{2^0}{2^3}=2^{-3}\)
\(\Rightarrow x=-3\)
a) \(4^x+4^{x+3}=4160\)
\(\Rightarrow4^x+4^x.4^3=4160\)
\(\Rightarrow4^x.\left(1+4^3\right)=4160\)
\(\Rightarrow4^x.65=4160\)
\(\Rightarrow4^x=64\)
\(\Rightarrow4^x=4^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
b) \(2^{x-1}+5.2^{x-2}=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{1}{2}+5.2^x.\frac{1}{4}=\frac{7}{32}\)
\(\Rightarrow2^x.\left(\frac{1}{2}+5.\frac{1}{4}\right)=\frac{7}{32}\)
\(\Rightarrow2^x.\frac{7}{4}=\frac{7}{32}\)
\(\Rightarrow2^x=\frac{7}{32}:\frac{7}{4}\)
\(\Rightarrow2^x=\frac{1}{8}\)
\(\Rightarrow2^x=2^{-3}\)
\(\Rightarrow x=-3\)
Vậy \(x=-3\)
\(2^{x-1}+5.2^{x-1}=\frac{7}{32}\)
=> \(2^{x-1}\left(1+5\right)=\frac{7}{32}\)
=> \(2^{x-1}.6=\frac{7}{32}\)
=> \(2^{x-1}=\frac{7}{32}:6=\frac{7}{192}\)
ĐỀ SAI RỒI BẠN !
a) 2(x + 2)2 - 100 = 62
=> 2(x + 2)2 = 62 + 100
=> 2(x + 2)2 = 162
=> (x + 2)2 = 162 : 2
=> (x + 2)2 = 81
mà \(9^2=81\) và \(\left(-9\right)^2=81\)
nên x + 2 = 9 và x + 2 = -9
* x + 2 = 9 => x = 9 - 2 = 7
* x + 2 = -9 => x = -9 - 2 = -11
b) 5 . 2x - 2x + 1 = 48
=> 5 . 2x - 2x . 2 = 48
=> 2x (5 - 2) = 48
=> 2x . 3 = 48
=> 2x = 48 : 3
=> 2x = 16
=> x = 4
\(a)2(x+2)^2-100\)\(=62\)
suy ra \(2(x+2)^2 =62+100\)
suy ra \(2(x+2)^2=162\)
suy ra \((x+2)^2=162/2\)
suy ra \((x+2)^2=81\)
suy ra\(x+2=9 hay x+2=-9\)
suy ra x=9-2=7 hay x=-9-2=-11
b)5.2^x-2x+1=48
suy ra \(5.2^x-2^x.2=48\)
suy ra \(2^x.(5-2)=48\)
suy ra \(2^x.3=48\)
suy ra \(2^x=48/3=16\)
suy ra\(2^x=2^4\)
suy ra \(x=4\)
\(5\cdot2^{x+1}\cdot2^{-2}-2^x=384\\ 5\cdot\frac{1}{4}\cdot2^x\cdot2-2^x=384\\ 2^x\left(\frac{5}{2}-1\right)=384\\ 2^x\cdot\frac{3}{2}=384\\ 2^x=\frac{384}{\frac{3}{2}}\\ 2^x=384\cdot\frac{2}{3}=256=2^8\\ \Rightarrow x=8\)
Vậy x = 8