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

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

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

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

=> không có giá trị x,y,z thỏa mãn đề

\(\Leftrightarrow x+y+z=2\sqrt{x-2}+2\sqrt{y+2003}+2\sqrt{z-2004}\)

\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2003-2\sqrt{y+2003}+1\right)\)

\(+\left(z-2004-2\sqrt{z-2004}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2003}-1\right)^2+\left(\sqrt{z-2004}-1\right)^2=0\)

Vì biểu thức trên là tổng của các số hạng không âm nên nó bằng 0 khi và chỉ khi các số hạng phải bằng 0

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-2003}=1\\\sqrt{z-2004}=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=3\\y=2004\\z=2005\end{cases}}}\)

\(ĐK:x\ge2,y\ge-2003,z\ge2004\)

Pt đã cho tương đương :

\(x+y+z-2\sqrt{x-2}-2\sqrt{y+2003}-2\sqrt{z-2004}=0\)

\(\Leftrightarrow\left(x-2-2\sqrt{x-2}+1\right)+\left(y+2003-2\sqrt{y+2003}+1\right)+\left(z-2004-2\sqrt{z-2004}+1\right)\)\(=0\)

\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2003}-1\right)^2+\left(\sqrt{z-2004}-1\right)^2=0\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x-2=1\\y+2003=1\\z-2004=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=-2002\\z=2005\end{cases}}\)(Thỏa mãn)

\(\dfrac{x-4}{2001}\)- 1 +\(\dfrac{x-3}{2002}\)-1 + \(\dfrac{x-2}{2003}\)-1 =\(\dfrac{x-2003}{2}\)-1 + \(\dfrac{x-2002}{3}\)-1 +\(\dfrac{x-2001}{4}\)-1 <=> \(\dfrac{x-2005}{2001}\)+\(\dfrac{x-2005}{2002}\)+\(\dfrac{x-2005}{2003}\)-\(\dfrac{x-2005}{2}\)-\(\dfrac{x-2005}{3}\)-\(\dfrac{x-2005}{4}\)= 0 <=> (x-2005). (\(\dfrac{1}{2001}\)+\(\dfrac{1}{2002}\)+\(\dfrac{1}{2003}\)-\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)) =0 <=> x-2005=0 ( vì \(\dfrac{1}{2001}\) +\(\dfrac{1}{2002}\) +\(\dfrac{1}{2003}\)- \(\dfrac{1}{2}\) -\(\dfrac{1}{3}\)- \(\dfrac{1}{4}\) khác 0) =>x = 2005

x-4/2001+ x-3/2002 + x-2/2003= x-2003/2 + x-2002/3 + x-2001/4

<=>(x-4/2001 -1)+(x-3/2002 -1)+(x-2/2003 -1)-(x-2003/2 -1)+

(x-2002/3 -1)+(x-2001/4 -1) =0

<=>x-2005/2001+ x-2005/2002+ x-2005/2003- x-2005/2-

x-2005/3- x-2005/4 =0

<=>(x-2005).(1/2001+1/2002+1/2003- 1/2- 1/3- 1/4)=0

<=>x-2005=0 (vì 1/2001+1/2002+1/2003-1/2-1/3-1/4)

<=>x=2005

Vậy pt có nghiệm là x=2005

\(\dfrac{x-4}{2001}+\dfrac{x-3}{2002}+\dfrac{x-2}{2003}=\dfrac{x-2003}{2}+\dfrac{x-2002}{3}+\dfrac{x-2001}{4}\)

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

\(\Leftrightarrow\dfrac{x-2005}{2001}+\dfrac{x-2005}{2002}+\dfrac{x-2005}{2003}-\dfrac{x-2005}{2}-\dfrac{x-2005}{3}-\dfrac{x-2005}{4}=0\)

\(\Leftrightarrow\left(x-2005\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}\ne0\right)=0\)

\(\Leftrightarrow x-2005=0\)

\(\Leftrightarrow x=2005\)

Vậy nghiệm của PT là \(x=2005\)

\(\frac{x}{2000}+\frac{x+1}{2001}+\frac{x+2}{2002}+\frac{x+3}{2003}=4\)

\(\Leftrightarrow\left(\frac{x}{2000}-1\right)+\left(\frac{x+1}{2001}-1\right)+\left(\frac{x+2}{2002}-1\right)+\left(\frac{x+3}{2003}-1\right)=4-4=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x-2000=0\)  ( do \(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\ne0\) )

\(\Leftrightarrow x=2000\)

Vậy x = 2000

Đây là cách của lớp 7 nha

@@ Học tốt

\(\frac{x}{2000}\)- 1+\(\frac{x+1}{2001}\)-1+\(\frac{x+2}{2002}\)-1+\(\frac{x+3}{2003}\)-1=0

<=>\(\frac{x-2000}{2000}\)+ \(\frac{x-2000}{2001}\)+ \(\frac{x-2000}{2002}\)+ \(\frac{x-2000}{2003}\)=0

<=>\(\left(x-2000\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)=0

Do \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\)khác 0

=> \(x-2000=0\)<=> \(x=2000\)

\(\Leftrightarrow\frac{x}{2000}-1+\frac{x+1}{2001}-1+\frac{x+2}{2002}-1+\frac{x+3}{2003}-1=0\)

\(\Leftrightarrow\frac{x-2000}{2000}+\frac{x-2000}{2001}+\frac{x-2000}{2002}+\frac{x-2000}{2003}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)

\(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}>0\)

\(\Rightarrow x-2000=0\)\(\Rightarrow x=2000\)