h) \(=\frac{x+1}{32}\)

câu hỏi tương tự

\(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{4}\right)+\left(x+\dfrac{1}{8}\right)+\left(x+\dfrac{1}{16}\right)=1\)

\(\Leftrightarrow4x-\dfrac{1}{16}=0\)

\(\Leftrightarrow x=\dfrac{1}{64}\)

Hộ e giải bài này vs ạ

( x + 1/2 ) + ( x + 1/4 ) + ( x + 1/8 ) + ( x + 1/16 ) = 1

( x + x + x + x ) + ( 1/2 + 1/4 + 1/8 + 1/16 ) = 1

x * 4 + 15/16 = 1

x * 4 = 1 - 15/16

x * 4 = 1/16

x = 1/16 : 4

x = 1/16 x 1/4

x = 1/64

x+x+x+x+1/2+1/4+1/8+1/16 = 1

x.4 + 15/16 = 1

x.4 = 1 – 15/16 = 1/16

x = 1/16 : 4

x = 1/64

a) 2^x . 16^2 = 1024                          b) 64 . 4^x =  16^8                    c) 2^x = 16

=> 2^x . 256 = 1024                             => 64 . 4^x  = (4^2) ^ 8               => 2^x  = 2^4

=> 2^x          = 1024 : 256                    =>   4^3 . 4^x  =  4^16                  => x     = 4

=> 2^x           =  4                                 =>             4^x  = 4^16 : 4^3

=>  2^x           = 2^2                             =>              4^x  =  4^13

                                                              =>                 x   =  13 

=>      x           = 2

a) \(2^x.16^2=1024\Rightarrow2^x=1024:16^2=2^{10}:\left(2^4\right)^2=2^{10}:2^8=2^2\)\(\Rightarrow x=2\)

b) \(64.4^x=16^8\Rightarrow4^x=16^8:64=\left(4^2\right)^8:4^3=4^{16}:4^3=4^{13}\Rightarrow x=13\)

c)\(2^x=16\Rightarrow2^x=2^4\Rightarrow x=4\)

1/2 nha bn

\(\frac{1}{8}-\frac{1}{8}\cdot x=\frac{1}{16}\)

\(\frac{1}{8}\left(1-x\right)=\frac{1}{16}\)

\(1-x=\frac{1}{16}:\frac{1}{8}\)

\(1-x=\frac{1}{2}\)

\(x=1-\frac{1}{2}\)

\(x=\frac{1}{2}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)

\(4x+\left(1-\frac{1}{16}\right)=1\)

\(4x+\frac{15}{16}=1\)

\(4x=\frac{1}{16}\)

\(\Rightarrow x=\frac{1}{64}\)

\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)

\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)

\(4x+\frac{15}{16}=1\)

\(4x=\frac{1}{16}\)

\(x=\frac{1}{16}\div4\)

\(x=\frac{1}{64}\)

Vậy ...

\(a,x-\dfrac{7}{12}x=\dfrac{5}{24}-\dfrac{3}{8}x\)

\(\Leftrightarrow\dfrac{5}{12}x+\dfrac{3}{8}x=\dfrac{5}{24}\)

\(\Leftrightarrow\dfrac{19}{24}x=\dfrac{5}{24}\Leftrightarrow x=\dfrac{5}{19}\)

Vậy x = 5/19

\(b,\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=0\\-3-\dfrac{x}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-6\end{matrix}\right.\)

Vậy x = 1/2 hoặc x = -6

\(c,\dfrac{x-3}{-2}=\dfrac{-8}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=4\\x-3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-1\end{matrix}\right.\)

Vậy x = 7 hoặc x = -1