\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+1+\dfrac{x+6}{2004}+1+\dfrac{x+7}{2003}+1=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

b) \(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}+\dfrac{3}{97.100}\\ =1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}=\dfrac{99}{100}\)

\(B=\left(2x-1\right)^2+\left(x+2\right)^2\\ =4x^2-4x+1+x^2+4x+4\\ =5x^2+5\)

\(\forall x\inℝ:x^2\ge0\Rightarrow5x^2\ge0\\ \Rightarrow B=5x^2+5\ge5\\ \Rightarrow Min_B=5\)

Dấu = xảy ra khi: \(x^2=0\Leftrightarrow x=0\)

\(\dfrac{4}{10000}\)

\(B=4x^2-4x+1+x^2+4x+4\\ =5x^2+5\ge5\forall x\inℝ\) (Vì: \(\forall x\inℝ\Rightarrow x^2\ge0\))

Dấu = xảy ra khi: \(x^2=0\Leftrightarrow x=0\)

\(Min_B=5tạix=0\)

\(A=x^2-3x+5\\ =x^2-2.x.\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2-\dfrac{9}{4}+5\\ =\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\inℝ\)

Dấu = xảy ra khi: \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)

\(Min_A=\dfrac{11}{4}tạix=\dfrac{3}{2}\)

\(B=-x^2-2y^2+2xy-2x+10y-40\\ =-\left(x^2-2xy+y^2\right)-y^2-2x+10y-40\\ =-\left(x-y\right)^2-2\left(x-y\right)-1-y^2+8y-39\\ =-\left[\left(x-y\right)^2+2.\left(x-y\right).1+1^2\right]-\left(y^2-8y+16\right)-23\\ =-\left(x-y+1\right)^2-\left(y-4\right)^2-23\le-23< 0\left(DPCM\right)\)

Dấu = xảy ra khi: \(x-y+1=y-4=0< \\ \Leftrightarrow y=4,x=3\)

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

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

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

Vậy: \(x\in\left\{-3;-5\right\}\)

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

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