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3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
\(1,\Rightarrow3^{x-3}=\left(3^2\right)^8:\left(3^3\right)^5=3^{16}:3^{15}=3^1\\ \Rightarrow x-3=1\\ \Rightarrow x=4\\ 2,\Rightarrow7^x\left(1+7^2\right)=350\\ \Rightarrow7^x=\dfrac{350}{50}=7=7^1\\ \Rightarrow x=1\)
\(3,\Rightarrow2^{2+2x+2}-2^{2x}=240\\ \Rightarrow2^{2x}\left(2^4-1\right)=240\\ \Rightarrow2^{2x}=\dfrac{240}{15}=16=2^4\\ \Rightarrow2x=4\Rightarrow x=2\)
Đặt \(A=2^{2x}+2^{2x+1}+...+2^{2x+1918}\)
=>\(2\cdot A=2^{2x+1}+2^{2x+2}+...+2^{2x+1919}\)
=>\(A=2^{2x+1919}-2^{2x}\)
Theo đề, ta có; \(2^{2x+1919}-2^{2x}=2^{1923}-2^4\)
=>\(2^{2x}\cdot\left(2^{2019}-1\right)=2^4\left(2^{2019}-1\right)\)
=>2x=4
=>x=2
(4x - 2008 ) : 4 - 48 = 48
( 4x - 2008 ) : 4 = 48 +48
(4x - 2008 ) : 4 = 96
4x - 2008 = 96.4
4x - 2008 = 384
4x = 384 + 2008
4x = 2382
x = 598
Tk mk nha
\(2^{2x+1}-15=17\\ =>2^{2x+1}=32=2^5\\ =>2x+1=5\\ =>2x=4\\ =>x=2\)
22x + 1 - 15 = 17
22x+1 = 17 + 15
22x+1 = 32
22x+1 = 25
2x + 1 = 5
2x = 5 - 1
2x = 4
x = 4 : 2
x = 2
Vậy x = 2.
a: \(\Leftrightarrow\left(\dfrac{13}{4}:x\right)\cdot\left(-\dfrac{5}{4}\right)=\dfrac{-10}{6}-\dfrac{5}{6}=\dfrac{-15}{6}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{13}{4}:x=\dfrac{5}{2}\cdot\dfrac{5}{4}=\dfrac{25}{8}\)
hay \(x=\dfrac{13}{4}:\dfrac{25}{8}=\dfrac{13}{4}\cdot\dfrac{8}{25}=\dfrac{26}{25}\)
b: \(\Leftrightarrow\dfrac{3}{4}:x=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{2}{36}=\dfrac{1}{18}\)
=>\(x=\dfrac{3}{4}:\dfrac{1}{18}=\dfrac{54}{4}=\dfrac{27}{2}\)
c: \(\Leftrightarrow\left(-\dfrac{6}{5}+x\right):\left(-3.6\right)=-\dfrac{7}{4}+\dfrac{1}{4}\cdot8=\dfrac{1}{4}\)
=>x-6/5=-9/10
=>x=3/10
Ta có : (3x-7)2-1=48
(3x-7)2 = 48 + 1 = 49
TH 1 : 3x - 7 = 7 => 3x = 7 + 7 = 14 => x = 14 : 3 = \(\frac{14}{3}\)
TH 2 : 3x - 7 = -7 => 3x = -7 + 7 = 0 => x = 0 : 3 = 0
(4x-2008):4-48=48
(4x-2008):4=48+48
(4x-2008):4=96
4x-2008 =96.4
4x-2008 =384
4x =384+2008
4x =2392
x =2392:4
x =598
Vậy x = 598
\(\left(4x-2008\right):4-48=48\)
\(\Rightarrow\left(4x-2008\right):4=48+48\)
\(\Rightarrow\left(4x-2008\right):4=96\)
\(\Rightarrow\left(4x-2008\right)=96.4\)
\(\Rightarrow\left(4x-2008\right)=384\)
\(\Rightarrow4x=384+2008\)
\(\Rightarrow4x=2392\)
\(\Rightarrow x=2396:4\)
\(\Rightarrow x=598\)
[( 4x + 48 ) . 3 + 55 ] : 5 = 35
( 4x + 48 ) . 3 + 55 = 35 . 5 = 175
4x + 48 = 175 : 3 = 175/3
4x = 175/3 - 48 = 31/3
x = 31/3 : 4 = 31/12=2/7/12
\(4^x+2^{2x+1}=48\\ \left(2^2\right)^x+2^{2x}\cdot2=48\\ 2^{2x}\cdot\left(1+2\right)=48\\ 2^{2x}\cdot3=48\\ 2^{2x}=48:3\\ 2^{2x}=16\\ 2^{2x}=2^4\\ 2x=4\\ x=4:2\\ x=2\)
4x - 22x+1 = 48
4x - 4x+1 = 48
4x - 4x . 4 = 48
4x . ( 4 - 1 ) = 48
4x . 3 = 48
4x = 48 : 3
4x = 16
4x = 42
x = 2
Vậy x = 2