Bài giải:

Điều kiện: $x\ne 0,x\ne -1,x\ne \frac{1}{2}$

$M=\left( \frac{x+2}{3x}+\frac{2}{x+1}-3 \right):\frac{2-4x}{x+1}-\frac{3x-{{x}^{2}}+1}{3x}$

=$\frac{\left( x+2 \right)\left( x+1 \right)+2.3x-3.3x.\left( x+1 \right)}{3x\left( x+1 \right)}.\frac{x+1}{2\left( 1-2x \right)}-\frac{3x-{{x}^{2}}+1}{3x}$

=$\frac{2\left( 1-2x \right)\left( 1+2x \right)}{3x\left( x+1 \right)}.\frac{x+1}{2.\left( 1-2x \right)}-\frac{3x-{{x}^{2}}+1}{3x}$

=$\frac{1+2x}{3x}-\frac{3x-{{x}^{2}}+1}{3x}$

=$\frac{{{x}^{2}}-x}{3x}$

=$\frac{x-1}{3}$

Khi x=5947 thì $M=\frac{5947-1}{3}=1982$

Đáp án đúng là D