10982 x 1024
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Tìm x: \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16} +...-\dfrac{1}{1024}=\dfrac{x}{1024}\)
\(\dfrac{x}{1024}=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+...-\dfrac{1}{1024}\)
\(\dfrac{2x}{1024}=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+...-\dfrac{1}{512}\)
\(\Rightarrow\dfrac{x}{1024}+\dfrac{2x}{1024}=1-\dfrac{1}{1024}\)
\(\Rightarrow\dfrac{3x}{1024}=\dfrac{1023}{1024}\)
\(\Rightarrow3x=1023\)
\(\Rightarrow x=341\)
Lời giải:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
\(\left(x+1\right)^{2y}=1024\)
\(\Rightarrow\left[\left(x+1\right)^y\right]^2=32^2\)
\(\Rightarrow\left(x+1\right)^y=32\)
Do \(x,y\in Z\)
\(\Rightarrow\left\{{}\begin{matrix}x+1=2\\y=5\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
\(2^x+2^{x+2}+2^{x+4}=1024\)
=>\(2^x\left(1+2^2+2^4\right)=1024\)
=>\(2^x=\dfrac{1024}{21}\)
=>\(x=log_2\left(\dfrac{1024}{21}\right)\)
256 x 6=512 x 3
512 x 3=256 x 6
999 x 2>1000
512 x 5>1024
1024<512 x 5
1024=512 x 2
512=256 x 2
256=128 x 2
128=64 x 2
64=32 x 2
32=16 x 2
\(2^x\cdot16^2=1024\)
\(\Leftrightarrow2^x=4\)
hay x=2
\(2^ax2^b=2^{\left(a+b\right)}=2014=2^{10}\)
\(\Rightarrow a+b=10\)
=11245568
11245568 nha bạn
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