\(\dfrac{x}{2000}+\dfrac{x+1}{2001}+\dfrac{x+2}{2002}+\dfrac{x+3}{2003}+\dfrac{x+4}{2004}=5\)

\(\Leftrightarrow\dfrac{x}{2000}-1+\dfrac{x+1}{2001}-1+\dfrac{x+2}{2002}-1+\dfrac{x+3}{2003}-1+\dfrac{x+4}{2004}-1=0\)

\(\Leftrightarrow\dfrac{x-2000}{2000}+\dfrac{x-2000}{2001}+\dfrac{x-2000}{2002}+\dfrac{x-2000}{2003}+\dfrac{x-2000}{2004}=0\)

\(\Leftrightarrow\left(x-2000\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)

Mà \(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}>0\)

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

Vậy x = 2000

`(2-x)/2002-1=(1-x)/2003-x/2004`

`<=>(2-x)/2002-1+(x-1)/2003+x/2004=0`(chuyển  vế)

`<=>(2-x)/2002+1+(x-1)/2003-1+x/2004-1=0`

`<=>(2004-x)/2002+(x-2004)/2003+(x-2004)/2004=0`

`<=>(x-2004)(1/2003+1/2004-1/2002)=0`

`<=>x=2004` do `1/2003+1/2004-1/2002 ne 0`

Vậy `x=2004`

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

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)=0\)

\(\Leftrightarrow x=-2005\)

1.

\(\dfrac{x+1}{x-2}=\dfrac{1}{x^2-4}\),\(ĐKXĐ,x\ne\pm2\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{\left(x-2\right)\left(x+2\right)}\)

\(\Leftrightarrow x^2+3x+2=1\)

\(\Leftrightarrow x^2+3x+1=0\)

\(\Leftrightarrow\dfrac{-3+\sqrt{9-4.1.1}}{2}=\dfrac{-3+\sqrt{5}}{2}\left(TM\right)\)

Vậy nghiệm PT là: \(x=\dfrac{-3+\sqrt{5}}{2}\)

b.

ta có : \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

\(\Leftrightarrow-\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)

\(\Leftrightarrow\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)

\(\Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)

\(\Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)

Vì \(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\Rightarrow2004-x=0=>x=2014\)

Vậy nghiệm của PT là x = 2014

\(\dfrac{x+101}{2001}+\dfrac{x+99}{2003}=\dfrac{x+100}{2002}+\dfrac{x+98}{2004}\)

\(\Leftrightarrow\left(\dfrac{x+101}{2001}+1\right)+\left(\dfrac{x+99}{2003}+1\right)=\left(\dfrac{x+100}{2002}+1\right)+\left(\dfrac{x+98}{2004}+1\right)\)

\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}=\dfrac{x+2102}{2002}+\dfrac{x+2102}{2004}\)

\(\Leftrightarrow\dfrac{x+2102}{2001}+\dfrac{x+2102}{2003}-\dfrac{x+2102}{2002}-\dfrac{x+2102}{2004}=0\)

\(\Leftrightarrow\left(x+2102\right)\left(\dfrac{1}{2001}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\right)=0\)

Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2004}\ne0\)

\(\Rightarrow x+2102=0\)

\(\Rightarrow x=-2102\)

\(\Rightarrow S=\left\{-2102\right\}\)

a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)

\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)

b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)

a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền

b,  - 1 hoặc + 1 vs mỗi phân số nha

BPT \(\Leftrightarrow\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}>0\)

\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)>0\)

\(\Leftrightarrow\dfrac{x-15}{2002}+\dfrac{x-15}{2003}-\dfrac{x-15}{2004}-\dfrac{x-15}{2005}>0\)

\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)

Vì \(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}>0\)

\(\Rightarrow x-15>0\)

\(\Leftrightarrow x>15\)

Vậy bpt có nghiệm x > 15

\(\dfrac{x+1987}{2002}+\dfrac{x+1988}{2003}-2>\dfrac{x+1989}{2004}+\dfrac{x+1990}{2005}-2\)

\(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)\)

\(-\left(\dfrac{x+1989}{2004}-1\right)-\left(\dfrac{x+1990}{2005}-1\right)\)

quy đồng lên ta được:

\(\left(\dfrac{x+1987-2002}{2002}\right)+\left(\dfrac{x-1998-2003}{2003}\right)\)

\(-\left(\dfrac{x+1989-2004}{2004}\right)-\left(\dfrac{x+1990-2005}{2005}\right)>0\)

\(\Leftrightarrow\left(\dfrac{x-15}{2002}\right)+\left(\dfrac{x-15}{2003}\right)-\left(\dfrac{x-15}{2004}\right)-\left(\dfrac{x-15}{2005}\right)>0\)

đặt nhân tử chung ta được:

\(\Leftrightarrow\left(x-15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)>0\)

Vì:

\(\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\in Z\right)\) nên ta xét \(x-15>0\Rightarrow x>15\)

Kí hiệu bất phương trình là (*). Ta có:

(*) \(\Leftrightarrow\left(\dfrac{x+1987}{2002}-1\right)+\left(\dfrac{x+1988}{2003}-1\right)>\left(\dfrac{x+1989}{2004}-1\right)+\left(\dfrac{x+1990}{2005}-1\right)\\ \Leftrightarrow\dfrac{x-15}{2002}+\dfrac{x-15}{2003}-\dfrac{x-15}{2004}-\dfrac{x-15}{2005}>0\\ \Leftrightarrow\left(x+15\right)\left(\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}\right)\)

Ta có:

\(\dfrac{1}{2002}>\dfrac{1}{2004}\Rightarrow\dfrac{1}{2002}-\dfrac{1}{2004}>0\\ \dfrac{1}{2003}>\dfrac{1}{2005}>\dfrac{1}{2003}-\dfrac{1}{2005}>0\\ \Rightarrow\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}>0\)

\(\Rightarrow\)(*) \(\Leftrightarrow x-15>0\Leftrightarrow x>15\)

Tập nghiệm \(S=\left\{x|x>15\right\}\)

\(\text{a) }\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\\ \Leftrightarrow\dfrac{2-x-2002}{2002}=\left(\dfrac{1-x}{2003}-1\right)+\left(1-\dfrac{x}{2004}\right)\\ \Leftrightarrow\dfrac{2004-x}{2002}-\dfrac{2003-x}{2003}-\dfrac{2004-x}{2004}=0\\ \Leftrightarrow\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\\ \Leftrightarrow2004-x=0\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\right)\\ \Leftrightarrow x=2004\)

Vậy phương trình có nghiệm \(x=2004\)

\(\text{b) }\dfrac{x^2-10x-29}{1971}+\dfrac{x^2-10x-27}{1973}=\dfrac{x^2-10x-1971}{29}+\dfrac{x^2-10x-1973}{27}\left(\text{ Chữa đề }\right)\\ \Leftrightarrow\left(\dfrac{x^2-10x-29}{1971}-1\right)+\left(\dfrac{x^2-10x-27}{1973}-1\right)=\left(\dfrac{x^2-10x-1971}{29}-1\right)+\left(\dfrac{x^2-10x-1973}{27}-1\right)\\ \Leftrightarrow\dfrac{x^2-10x-2000}{1971}+\dfrac{x^2-10x-2000}{1973}-\dfrac{x^2-10x-2000}{29}-\dfrac{x^2-10x-2000}{27}=0\\ \Leftrightarrow\left(x^2-10x-2000\right)\left(\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\right)=0\\ \Leftrightarrow x^2-10x-2000=0\left(\text{Vì }\dfrac{1}{1971}+\dfrac{1}{1973}-\dfrac{1}{29}-\dfrac{1}{27}\ne0\right)\\ \Leftrightarrow x^2-20x+10x-2000=0\\ \Leftrightarrow x\left(x-20\right)+10\left(x-20\right)=0\\ \Leftrightarrow\left(x+10\right)\left(x-20\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+10=0\\x-20=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-10\\x=20\end{matrix}\right.\)

Vậy phương trình có tập nghiệm \(S=\left\{-10;20\right\}\)