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\(\Leftrightarrow\dfrac{5}{2}\cdot2\cdot2^x-2^x=384\)

\(\Leftrightarrow4\cdot2^x=384\)

\(\Leftrightarrow2^x=96\)

hay \(x\in\varnothing\)

10 tháng 11 2019

\(5\cdot2^{x+1}\cdot2^{-2}-2^x=384\\ 5\cdot\frac{1}{4}\cdot2^x\cdot2-2^x=384\\ 2^x\left(\frac{5}{2}-1\right)=384\\ 2^x\cdot\frac{3}{2}=384\\ 2^x=\frac{384}{\frac{3}{2}}\\ 2^x=384\cdot\frac{2}{3}=256=2^8\\ \Rightarrow x=8\)

Vậy x = 8

14 tháng 10 2017

d.

33 < 3x < 35

--> 3 < x < 5

suy ra x=4

12 tháng 10 2017

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) \(3^{x+2}.5^y=45^x\Leftrightarrow3^{x+2}.5^y=3^{2x}.5^x\Leftrightarrow\frac{3^{2x}}{3^{x+2}}=\frac{5^y}{5^x}\)\(\Leftrightarrow3^{2x-x+2}=5^{y-x}\Leftrightarrow3^{x+2}=5^{y-x}\)

\(\Rightarrow x+2=y-x=0\Rightarrow x=y=-2\)

\(5.2^{x+1}.2^{-2}-2x=384\)

\(\Leftrightarrow5.2^x.2.2^{-2}-2^x=384\)

\(\Leftrightarrow5.2^x.2^{-1}-2^x=384\)

\(\Leftrightarrow2^x.\frac{5}{2}-2^x=384\)

\(\Leftrightarrow2^x\left(\frac{5}{2}-1\right)=384\)

\(\Leftrightarrow2^x\frac{3}{2}=284\)

\(\Leftrightarrow2^x=2^8\)

\(\Leftrightarrow x=8\)

30 tháng 10 2022

a: \(\Leftrightarrow2x-3=x\)

=>x=3

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)

=>2^x=1/8

=>x=-3

c: =>2x+7=-4

=>2x=-11

=>x=-11/2

d: =>(4x-3)^2*(4x-4)(4x-2)=0

hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)

16 tháng 9 2018

a, đầu bài sai hay sao đó bn

b,\(\dfrac{2}{3}\).3x+1 - 7.3x = -405

3x(\(\dfrac{2}{3}\).3 - 7) = -405

3x . (-5) = -405

3x = 81 mà 81 = 34 suy ra x = 4

15 tháng 9 2019

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: \(\dfrac{3x+2}{5x+7}=\dfrac{3x-1}{5x+1}\)

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

=>16x-7=13x+2

=>3x=9

hay x=3

b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)

=>x+2017=0

hay x=-2017

e: \(\left(2x-3\right)^2=144\)

=>2x-3=12 hoặc 2x-3=-12

=>2x=15 hoặc 2x=-9

=>x=15/2 hoặc x=-9/2