\(4^{x+3}+4^{x+2}+4^{x+1}+4^x=5440\)

\(\Rightarrow4^x.4^3+4^x.4^2+4^x.4+4^x=5440\)

\(\Rightarrow4^x\left(4^3+4^2+4+1\right)=5440\)

\(\Rightarrow4^x.\left(64+16+4+1\right)=5440\)

\(\Rightarrow4^x.85=5440\)

\(\Rightarrow4^x=5440:85\)

\(\Rightarrow4^x=64=4^3\)

\(\Rightarrow x=3\)

( x- 1/2) ²= 4

( x- 1/2) ²= 2²

=> x- 1/2 =2

     x         = 2+1/2

     x         = 5/2

Answer:

\(\left(x-\dfrac{1}{2}\right)^2=4\)

\(\left(x-\dfrac{1}{2}\right)^2=\left(\pm2\right)^2\)

TH1:  \(\left(x-\dfrac{1}{2}\right)^2=2^2\)

             \(x-\dfrac{1}{2}=2\)

             \(x=2+\dfrac{1}{2}\)

             \(x=\dfrac{5}{2}\)

TH2:  \(\left(x-\dfrac{1}{2}\right)^2=\left(-2\right)^2\)

             \(x-\dfrac{1}{2}=-2\)

             \(x=-2+\dfrac{1}{2}\)

             \(x=\dfrac{-3}{2}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

\(\left|x\right|=\frac{1}{5}-\frac{1}{4}=-\frac{1}{20}\)(vô lý)

Vậy không có x thỏa mãn đề 

\(\left|x\right|=\frac{1}{5}-\frac{1}{4}\)

\(\left|x\right|=-\frac{1}{20}\)

\(\Rightarrow x\in\varnothing\)

\(\left|x\right|=\frac{1}{5}-\frac{1}{4}\)

\(\Rightarrow\left|x\right|=-\frac{1}{20}\)

Ta có : Vì \(\left|x\right|\ge0\)

mà \(-\frac{1}{20}< 0\)

\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\varnothing\)

3) |x(x-4)| = x

\(\Rightarrow\hept{\begin{cases}x\left(x-4\right)=x\\x\left(x-4\right)=-x\end{cases}}\Rightarrow\hept{\begin{cases}x-4=1\\x-4=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=5\\x=3\end{cases}}\)

Vậy x = 3 hoặc x = 5

trị của x + 1 cộng trị của x + 4 > hoặc = 0

=> 3x > or = 0

=> x > or = 0

=> x + 1 + x + 4 = 3x

=> 2x + 5 = 3x

=> x = 5