n chẵn nên  đặt \(n=2k\)

\(n^3+2012n=8k^3+2012\cdot2k\)

\(8k^3+4024k\)

\(=8\left(k-1\right)k\left(k+1\right)+4032k\)

Mà \(\left(k-1\right)k\left(k+1\right)⋮6\Rightarrow8\left(k-1\right)\left(k+1\right)k⋮48;4032k⋮48\)

\(\Rightarrowđpcm\)

a, \(A=\frac{4x^2\left(x-2\right)+3\left(x-2\right)}{2x\left(x-2\right)+x-2}\)

\(=\frac{\left(x-2\right)\left(4x^2+3\right)}{\left(x-2\right)\left(2x+1\right)}=\frac{4x^2+3}{2x-1}\left(ĐKXĐ:x\ne2;x\ne-\frac{1}{2}\right)\)

b, \(A\in Z\Leftrightarrow\frac{4x^2+3}{2x-1}\in Z\Leftrightarrow2x+1+\frac{4}{2x-1}\in Z\)

\(\Leftrightarrow\frac{4}{2x-1}\in Z\Leftrightarrow4⋮\left(2x-1\right)\)

\(\Rightarrow2x-1\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

Mà 2x - 1 là số lẻ nên \(2x-1\in\left\{-1;1\right\}\Rightarrow x\in\left\{0;1\right\}\) (thỏa mãn ĐKXĐ)

a,\(\Delta AFE\infty\Delta BFD\left(g.g\right)\)

b, \(\Delta CBE\infty\Delta CAD\left(g.g\right)\Rightarrow\frac{CB}{CA}=\frac{CE}{CD}\Rightarrow\frac{CB}{CE}=\frac{CA}{CD}\)

c, Tam giác CEB có CM là tia p/g của \(\widehat{ECB}\left(M\in EB\right)\left(gt\right)\Rightarrow\frac{CB}{CE}=\frac{MB}{ME}\)

\(\Delta CDA\) có CN là tia phân giác của \(\widehat{ACD}\left(gt\right)\Rightarrow\frac{CA}{CD}=\frac{AN}{ND}\)

Mà \(\frac{CB}{CE}=\frac{CA}{CD}\left(cmt\right)\Rightarrow\frac{MB}{ME}=\frac{AN}{ND}\Rightarrow AN.ME=MB.ND\)

\(P=\frac{3a+7+2a-b-7}{3a+7}-\frac{2b-7+b-2a+7}{2b-7}\)

mà 2a-b=7 hay b-2a=-7 nên ta có

\(P=1+\frac{7-7}{3a+7}-1-\frac{-7+7}{2b-7}=1+0-1-0=0\)

\(A=\frac{4x+2}{x^2+2}=\frac{x^2+4x+4-\left(x^2+2\right)}{x^2+2}=\frac{\left(x+2\right)^2}{x^2+2}-1\ge-1\)

Dấu bằng xảy ra khi \(x=-2\)