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\(2^{2x}+2^{2x+1}=48\)
\(2^{2x}+2^{2x}.2=48\)
\(2^{2x}.\left(1+2\right)=48\)
\(2^{2x}.3=48\)
\(2^{2x}=48:3\)
\(2^{2x}=16\)
\(2^{2x}=2^4\)
\(\Rightarrow2x=4\)
\(x=4:2\)
\(x=2\)
(2x2 + 1)(x-3)=0
\(\Rightarrow\orbr{\begin{cases}2x^2+1=0\\x-3=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x^2=-1\Rightarrow x^2=-\frac{1}{2}\left(vl\right)\\x=3\end{cases}}\)
Vậy x=3
48-(15-x)5=48
(15-x)5=48-48
(15-x)5=0
=> 15-x =0
x =15-0
x =15
Vậy x=15
\(\Leftrightarrow\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right).2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{2}-\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow2x=24\)
\(\Rightarrow x=12\)
\(2^{x+1}-1=63\\ 2^{x+1}=64\\ 2^{x+1}=2^6\\ =>X+1=6\\ =>x=5\)
\(\left(X-3\right)\cdot4^5=4^8\\ X-3=64\\ =>X=67\)
1e) Để \(\frac{2x-1}{x-3}\) nguyên thì \(2x-1⋮x-3\)
\(\Leftrightarrow2x-6+5⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+5⋮x-3\)
Do \(2\left(x-3\right)⋮x-3\) \(\Rightarrow5⋮x-3\)
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\)
Vậy:...................
3 x (2x-1)2=48
=> (2x-1)2=48:3
=>(2x-1)2=16
Mà: 42=16 ; (-4)2=16
=> (2x-1)2 =42 hoặc (2x-1)2= (-4)2
=> 2x-1=4 hoặc 2x-1= -4
=> x= 2,5 hoặc x= -2,5
3 x ( 2x - 1 )2 = 48
=> ( 2x - 1 )2 = 48 : 3
=> ( 2x - 1 )2 = 16
=> ( 2x - 1 )2 = 42
=> 2x - 1 = 4 hoặc 2x - 1 = -4
+) 2x - 1 = 4
=> 2x = 5
=> x = 5/2
+) 2x - 1 = -4
=> 2x = -3
=> x = -3/2
Vậy x = 5/2 hoặc x = -3/2
\(\left(2x-4\right)^{38}=\left(2x-4\right)^{48}\)
\(\Rightarrow\left(2x-4\right)^{38}-\left(2x-4\right)^{48}=0\)
\(\Rightarrow\left(2x-4\right)^{38}\left[1-\left(2x-4\right)^{10}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(2x-4\right)^{38}=0\\1-\left(2x-4\right)^{10}=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4=0\\\left(2x-4\right)^{10}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x-4=1\\2x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\\2x=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{2;\dfrac{5}{2};\dfrac{3}{2}\right\}.\)
#\(Toru\)
2x+2x+1=48
2x+2x.21= 48
2x..(21+1)= 48
2x= 48 : 3
2x=16
2x=24
=> x = 4
Vậy x = 4
2x + 2x+1 = 48
2x. ( 1 + 2) = 48
2x . 3 = 48
2x = 48: 3
2x = 16
2x = 24
x = 4