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\(\left(\frac{1}{2}+x\right)+\left(\frac{1}{4}+x\right)+\left(\frac{1}{8}+x\right)+\left(\frac{1}{16}+x\right)=1\)
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)+\left(x+x+x+x\right)=1\)
\(\frac{15}{16}+4x=1\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{64}\)
( x + x + x + x ) + ( \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\) ) = 1
x * 4 + \(\frac{15}{16}\) = 1
x * 4 = 1 - \(\frac{15}{16}\)
x * 4 = \(\frac{1}{16}\)
x = \(\frac{1}{16}\) : 4
x = \(\frac{1}{64}\)
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a) 4x=2x+1
=> (22)x=2x+1
=> 22x=2x+1
=> 2x=x+1
=>x=1
b) 16=(x-1)4
=> (x-1)4 =(-2)4;24
Xet (x-1)4=(-2)4
=> x-1=-2
=> x=-1
Xet (x-1)4=24
=> x-1=2
=> x=3
Vậy x=-1;3
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\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right)=1\)
\(4x+\left(1-\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(\Rightarrow x=\frac{1}{64}\)
\(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{4}\right)+\left(x+\frac{1}{8}\right)+\left(x+\frac{1}{16}\right)=1\)
\(\left(x+x+x+x\right)+\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right)=1\)
\(4x+\frac{15}{16}=1\)
\(4x=\frac{1}{16}\)
\(x=\frac{1}{16}\div4\)
\(x=\frac{1}{64}\)
Vậy ...
16/1/X-1 (ĐK : x khác 1). (=)4/(x-1) = 16(x-1). (=) 16(x-1)^2=4. (=) (x-1)^2 = 1/4. (=) x-1 = 1/2 hoặc x-1 = -1/2. (=) x=1,5(t/mãn) hoặc x=0
đúng ko
\(x\over 16\)=\(1\over 4\)
<=>4x=16x1
<=>x=\(16\over 4\)
<=>x=4.
Vậy x=4