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	3	sửa đáp án		Sửa 26-07-16 10:10 PM ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 1
=>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{2002} \right )$ - $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ - $\left ( \frac{x+2004}{2003} \right )$ = 0 =>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} -\frac{1}{2002} -\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right )$ $>$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$ =>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{20002} \right )$ + $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ + $\left ( \frac{x+2004}{2003} \right )$ = 0=>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} +\frac{1}{2002} +\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right )$ $>$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$ =>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{2002} \right )$ - $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ - $\left ( \frac{x+2004}{2003} \right )$ = 0 =>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} -\frac{1}{2002} -\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right )$ $>$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$

	2	sửa đáp án		Sửa 26-07-16 10:08 PM ❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ 1
=>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{20002} \right )$ + $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ + $\left ( \frac{x+2004}{2003} \right )$ = 0=>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} +\frac{1}{2002} +\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right )$ $>$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$ =>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{20002} \right )$ + $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ + $\left ( \frac{x+2004}{2003} \right )$ = 0=>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} +\frac{1}{2002} +\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right )$ $\geq$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$ =>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{20002} \right )$ + $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ + $\left ( \frac{x+2004}{2003} \right )$ = 0=>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} +\frac{1}{2002} +\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right )$ $>$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$

	1	tạo mới		Trả lời 26-07-16 10:05 PM Yêu Tatoo 1
=>$\left ( \frac{x+4}{2000}+1 \right )$ + $\left ( \frac{x+3}{2001}+ 1 \right )$ = $\left ( \frac{x+2}{2002} +1\right )$ + $\left ( \frac{x+1}{2003}+ 1 \right )$=>$\left ( \frac{x+4+ 2000}{2000} \right )$ + $\left ( \frac{x+3+2001}{2001} \right )$ - $\left ( \frac{x+2+2002}{20002} \right )$ + $\left ( \frac{x+1+2003}{2003} \right )$ = $0$ =>$\left ( \frac{x+2004}{2000} \right )$+$\left ( \frac{x+2004}{2001} \right )$ - $\left ( \frac{x+2004}{2002} \right )$ + $\left ( \frac{x+2004}{2003} \right )$ = 0=>$\left ( x+2004 \right )$ $\left ( \frac{1}{2000} +\frac{1}{2001} +\frac{1}{2002} +\frac{1}{2003}\right )$ =0=>Vì $\left ( \frac{1}{2000} +\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right )$ $\geq$$0$=>$x$+$2004$ = $0$ => $x$$=$ $-2004$

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