\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\\ \Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\\ 3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow2B=1-\dfrac{1}{3^{100}}< 1\\ \Rightarrow B< \dfrac{1}{2}< 1\left(DPCM\right)\)

\(\left(4x+12\right)\left(x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x+12=0\\x+5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}4x=-12\\x=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-5\right\}\)

\(\left(4x+1\right)⋮\left(x-2\right)\\ \Rightarrow\left[4\left(x-2\right)+9\right]⋮\left(x-2\right)\\ \Rightarrow9⋮\left(x-2\right)\\ \Rightarrow x-2\inƯ\left(9\right)=\left\{1;-1;3;-3;9;-9\right\}\\ \Rightarrow x\in\left\{3;1;5;-1;11;-7\right\}\\ \underrightarrow{x\in N}x\in\left\{3;1;5;11\right\}\)

\(\left(3x-2\right)⋮\left(x-1\right)\Rightarrow\left[3\left(x-1\right)+1\right]⋮\left(x-1\right)\\ \Rightarrow1⋮\left(x-1\right)\Rightarrow x-1\inƯ\left(1\right)=\left\{1;-1\right\}\\ \Rightarrow x\in\left\{2;0\right\}\left(TM\right)\)

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Em đăng kí nhận GP

Em đăng kí nhận quà khảo sát.

\(14-8x^2=\dfrac{3}{2}\\ =>8x^2=14-\dfrac{3}{2}\\ =>8x^2=\dfrac{25}{2}\\ =>x^2=\dfrac{25}{2}:8\\ =>x^2=\dfrac{25}{16}\\ =>x=\pm\dfrac{5}{4}\)

\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Rightarrow1+2^2=\dfrac{1}{cos^2\alpha}\\ \Rightarrow5=\dfrac{1}{cos^2\alpha}\Rightarrow cos^2\alpha=\dfrac{1}{5}\)

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