Tìm x : 23x+4=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:
$\frac{x}{1024}=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...-\frac{1}{1024}$
$\frac{2x}{1024}=1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+...-\frac{512}$
$\Rightarrow \frac{x}{1024}+\frac{2x}{1024}=1-\frac{1}{1024}$
$\frac{3x}{1024}=\frac{1023}{1024}$
$\Rightarrow 3x=1023$
$\Rightarrow x=341$
a) \(18-\left(2x+5\right)=9\)
\(2x+5=18-9\)
\(2x+5=9\)
\(2x=9-5\)
\(2x=4\)
\(x=2\)
a) \(18-\left(2x+5\right)=9\)
\(\Rightarrow2x+5=18-9=9\)
\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)
b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)
c) \(\left(3x+2\right)^2=64\)
\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)
d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Lời giải:
$f'(x)=5(\sin ^23x-4)'(\sin ^23x-4)^4=5.2.\sin 3x (\sin 3x)'.(\sin ^23x-4)^4$
$=30\sin 3x\cos 3x(\sin ^23x-4)^4$
$\Rightarrow k=30$
Ta có: 8x.4x=1024
(8.4)^x=1024
32^x=1024
32^x=32^2
x=2
Vậy x=2
\(2^{3x+4}=1024\)
Đổi 1024=\(2^{10}\)
\(=>2^{3x+4}\)=\(2^{10}\)
3x+4=\(2^{10}\)
3x=10-4
3x=6
x=6:3
x=2
tk m nhé
\(2^{3x+4}=1024\)
Đổi: \(1024=2^{10}\)
\(\Rightarrow3x+4=10\)
\(3x=10-4\)
\(3x=6\)
\(x=6:3\)
\(x=2\)