\(\frac{\sqrt{x-2002}}{x-2002}-\frac{1}{x-2002}+\frac{\sqrt{y-2003}}{y-2003}-\frac{1}{y-2003}+\frac{\sqrt{z-2004}}{z-2004}-\frac{1}{z-2004}=\frac{3}{4}\)

\(1-\frac{1}{x-2002}+1-\frac{1}{y-2003}+1-\frac{1}{z-2004}=\frac{3}{4}\)

\(3-\frac{1}{x-2002}-\frac{1}{y-2003}-\frac{1}{z-2004}=\frac{3}{4}\)

\(\frac{1}{x-2002}+\frac{1}{y-2003}+\frac{1}{z-2004}=3-\frac{3}{4}=\frac{9}{4}\)

=> không có giá trị x,y,z thỏa mãn đề

có (x+1)(x+3)(x+5)(x+7)+2004

=(x²+8x+7)(x²+8x+15)+2004

=[(x²+8x+1)+6][(x²+8x+1)+14]+2004

=(x²+8x+1)²+20(x²+8x+1)+84+2004

=(x²+8x+1)²+20(x²+8x+1)+2088

vì (x²+8x+1)² chia hết chox²+8x+1

   20(x²+8x+1) chia hết cho x²+8x+1

=>(x+1)(x+3)(x+5)(x+7)+2004 chia cho x²+8x+1 dư 2088

84 ở đâu ra vậy 

(x+1)(x+3)(x+5)(x+7) + 2004

= ( x² + 8x + 7 ) ( x² + 8x + 15 ) + 2004

đặt x² + 8x + 1 = a

\(\Rightarrow\)( a + 6 ) ( a + 14 ) + 2004

= a² + 20a + 84 + 2004

= a² + 20a + 2088

Ta thấy a² + 20a \(⋮\)x² + 8x + 1 

\(\Rightarrow\)(x+1)(x+3)(x+5)(x+7) + 2004 chia x² + 8x + 1 dư 2088

\(\left(x-\frac{1}{2004}\right)+\left(x-\frac{2}{2003}\right)-\left(x-\frac{3}{2002}\right)=x-\frac{4}{2001}\)

\(x-\frac{1}{2004}+x-\frac{2}{2003}-x+\frac{3}{2002}-x=-\frac{4}{2001}\)

\(x+x-x-x-\frac{1}{2004}-\frac{2}{2003}+\frac{3}{2002}=-\frac{4}{2001}\)

\(0x-\frac{1}{2004}-\frac{2}{2003}+\frac{3}{2002}=-\frac{4}{2001}\)

\(\Rightarrow\) Vô lý

Vậy \(x\in\phi\)

đúng ko zay

ko hieu de x-1 hay la xhi x-