(3x-1)2.4=44
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Câu C giải rồi
\(B=\dfrac{1}{5}+\dfrac{1}{20}+\dfrac{1}{44}+\dfrac{1}{77}+\dfrac{1}{119}+\dfrac{1}{170}+\dfrac{1}{230}+\dfrac{1}{299}\)
\(=2\left(\dfrac{1}{10}+\dfrac{1}{40}+\dfrac{1}{88}+\dfrac{1}{154}+\dfrac{1}{238}+\dfrac{1}{340}+\dfrac{1}{460}+\dfrac{1}{598}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+\dfrac{3}{14.17}+\dfrac{3}{17.20}+\dfrac{3}{20.23}+\dfrac{3}{23.26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{23}-\dfrac{1}{26}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{26}\right)=\dfrac{4}{13}\)
Lời giải:
Biến đổi: \(q(x)=9.81^x+15.25^x+2.8^x+8.64^x\)
Lại có:
\(\left\{\begin{matrix} 81\equiv 13\pmod {17}\rightarrow 81^k\equiv 13^k\pmod {17}\\ 25\equiv 8\pmod {17}\rightarrow 25^k\equiv 8^k\pmod {17}\\ 64\equiv 13\pmod {17}\rightarrow 64^k\equiv 13^k\pmod {17}\end{matrix}\right.\)
Do đó, \(q(x)\equiv 9.13^k+15.8^k+2.8^k+8.13^k\pmod {17}\)
\(\Leftrightarrow q(x)\equiv 17.13^k+17.8^k\equiv 0\pmod {17}\)
\(\Leftrightarrow q(x)\vdots 17\) (đpcm)
3x - 24 . 45 = 2 . 46 . \(\frac{1}{2013^0}\)
3x - 16 . 1024 = 2 . 4096 . 1
3x - 16 384 = 8192
3x = 8192 + 16 384
3x = 24 576
x = 24 576 : 3
x = 8192
Vậy x = 8192
\(3x-16\cdot1024=2\cdot4096.1\)
3x - 16384 = 8192
3x = 8192 + 16384
3x = 24576
x = 24576 : 3
x = 8192
\(\left(3x-1\right)^2=144\\\Rightarrow\left[{}\begin{matrix}\left(3x-1\right)^2=12^2\\\left(3x-1\right)^2=\left(-12\right)^2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x-1=12\\3x-1=-12\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}3x=13\\3x=-11\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{3}\\x=-\dfrac{11}{3}\end{matrix}\right.\)
(3x - 1)2.4=44
(3x-1)2 = 44 : 4
(3x - 1)2 = 11
3x - 1 = + - \(\sqrt{11}\)
x = (\(\sqrt{11}\) + 1) : 3 ; x = (- \(\sqrt{11}\) + 1 ) : 3