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\(x:\dfrac{1}{2}+x:\dfrac{1}{4}+x:\dfrac{1}{8}+...+x:\dfrac{1}{512}=511\)
\(\Rightarrow x\left(2+4+8+...+512\right)=511\)
\(\Rightarrow\dfrac{\left(512+2\right).255}{2}.x=511\)
\(\Rightarrow65535x=511\)
\(\Rightarrow x=\dfrac{511}{65535}\)
Vậy.................
\(x:\dfrac{1}{2}+x:\dfrac{1}{4}+x:\dfrac{1}{8}+...+x:\dfrac{1}{512}=511\)
\(\Rightarrow x.\left(2+4+8+...+512\right)=511\)
\(\Rightarrow\dfrac{\left(512+2\right).255}{2}.x=511\)
\(\Rightarrow65535x=511\)
\(\Rightarrow x=\dfrac{511}{65535}\)
Vậy \(x=\dfrac{511}{65535}\)
1: =>x=3/5-1/5=2/5
b: =>x/3=5/8+1/8=3/4
=>x=9/4
3: =>10/3x=3+1/4+6+3/4=10
=>x=10:10/3=3
a: =>6/x=x/24
=>x^2=144
=>x=12 hoặc x=-12
b: =>x(1-7/12+3/8)=5/24
=>x*19/24=5/24
=>x=5/24:19/24=5/19
c: =>(x-1/3)^2=1+3/4+1/2=9/4
=>x-1/3=3/2 hoặc x-1/3=-3/2
=>x=11/6 hoặc x=-7/6
d: =>(x-3)^2=16
=>x-3=4 hoặc x-3=-4
=>x=-1 hoặc x=7
e: =>9/x=-1/3
=>x=-27
f: =>x-1/2=0 hoặc -x/2-3=0
=>x=1/2 hoặc x=-6
\(x+\left|\dfrac{1}{2}-\dfrac{1}{3}\right|=\left|\dfrac{-2}{3}-\dfrac{1}{4}\right|\)
\(x+\left|\dfrac{1}{6}\right|=\left|\dfrac{-11}{12}\right|\)
\(x+\dfrac{1}{6}=\dfrac{11}{12}\)
\(x=\dfrac{11}{12}-\dfrac{1}{6}\)
\(x=\dfrac{3}{4}\)
Vậy ...
\(\dfrac{x-1}{2}=\dfrac{8}{x-1}\)
\(\Rightarrow\left(x-1\right)\left(x-1\right)=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=\pm4\)
\(\Rightarrow\left[{}\begin{matrix}x-1=4\Rightarrow x=5\\x-1=-4\Rightarrow x=-3\end{matrix}\right.\)
\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)
\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}=\dfrac{2018-x}{2015}+\dfrac{2018-x}{2014}\)
\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)
\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)
Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)
\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)
Vậy ....
\(x-\dfrac{1}{2}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}+\dfrac{1}{2}\)
\(x=\dfrac{5}{4}\)
\(x+\dfrac{7}{8}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{7}{8}\)
\(x=\dfrac{-1}{8}\)
\(\dfrac{1}{2}\cdot x-\dfrac{1}{4}=\dfrac{-1}{2}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{2}+\dfrac{1}{4}\)
\(\dfrac{1}{2}\cdot x=\dfrac{-1}{4}\)
\(x=\dfrac{-1}{4}\div\dfrac{1}{2}\)
\(x=\dfrac{-1}{2}\)
Câu D ko bt
\(x:\dfrac{1}{2}+x:\dfrac{1}{4}+x:\dfrac{1}{8}+...+x:\dfrac{1}{512}=511\\ 2x+4x+8x+..+512x=511\\ x\left(2+4+8+...+512\right)=511\\ x\left(2^1+2^2+2^3+...+2^9\right)=511\\ \)
Gọi \(S=2^1+2^2+2^3+...+2^9\)
\(2S=2^2+2^3+2^4+...+2^{10}\\ 2S-S=\left(2^2+2^3+2^4+...+2^{10}\right)-\left(2^1+2^2+2^3+...+2^9\right)\\ S=2^{10}-2\)
\(x\left(2^{10}-2\right)=511\\ 2x\left(2^9-1\right)=511\\ 2x\left(512-1\right)=511\\ 2x\cdot511=511\\ 2x=1\\ x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)