Ta có: \(\frac{x+2}{2005}+\frac{x+3}{2004}+\frac{x+4}{2003}+3=0\)

\(\Leftrightarrow\left(\frac{x+2}{2005}+1\right)+\left(\frac{x+3}{2004}+1\right)+\left(\frac{x+4}{2003}+1\right)=0\)

\(\Leftrightarrow\frac{x+2007}{2005}+\frac{x+2007}{2004}+\frac{x+2007}{2003}=0\)

\(\Leftrightarrow\left(x+2007\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

Mà \(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}>0\) \(\Rightarrow x+2007=0\Rightarrow x=-2007\)

Hợp chất với H là :  XH4 \(\rightarrow\)%H=\(\frac{4}{X+4}\)

Oxit là XO2 \(\rightarrow\) \(\%O=\frac{32}{X+32}\)

64%H = 15%O \(\rightarrow\frac{64.4}{4+X}=\frac{15.32}{32+X}\)

\(\rightarrow X=28\)

Vậy X là nguên tố SI

mik cần gấp

đó bài giải nè

\(\left(2x^2-x+2\right)\left(2x^2-3x-2\right)-8x^2=0\)

\(\Leftrightarrow4x^4-6x^3-4x^2-2x^3+3x^2+2x+4x^2-6x-4-8x^2=0\)

\(\Leftrightarrow4x^4-8x^3-5x^2-4x-4=0\)

x^5 - 25x + 1 x + 2 x^4 - 2x^3 + 4x^2 - 8x - 9 x^5 + 2x^4 -2x^4 - 25x -2x^4 - 4x^3 4x^3 - 25x 4x^3 + 8x^2 -8x^2 - 25x -8x^2 - 16x -9x + 1 -9x - 18 19

Vậy \(\left(x^5-25x+1\right):\left(x+2\right)=x^4-2x^3+4x^2-8x-9\)dư 19