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\(\dfrac{2001+2002}{2002+2003}< \dfrac{2001}{2002}+\dfrac{2002}{2003}\)

Ta có

Ta có: \(\dfrac{2002}{2003}< \dfrac{14}{13}\) vì \(\dfrac{2002}{2003}< 1\), \(\dfrac{14}{13}>1\)

ta có

2002/2003<1<14/13

=>2002/2003<14/13

2001/2002=1-1/2002

2002/2003=1-1/2003

vi 1/2003<1/2002 nen 2001/2002<2002/2003

Ta có: 2003 x 2001 < 2002 x 2002

=> \(\frac{2001}{2002}\)<\(\frac{2002}{2003}\)

2 phân số trên: 2001/2002 = 2002/2003

\(\dfrac{x-4}{2001}\)- 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 \(x=2005\)

Ta có: \(2003^{2003}+1=2003^{2002+1}+1và2003^{2004}+1=2003^{2003+1}+1\)

\(\Rightarrow A>B\)

A > B

\(a,\dfrac{199}{200}=1-\dfrac{1}{200};\dfrac{200}{201}=1-\dfrac{1}{201}\\ Vì:\dfrac{1}{200}>\dfrac{1}{201}\\ \Rightarrow1-\dfrac{1}{200}< 1-\dfrac{1}{201}\\ Vậy:\dfrac{199}{200}< \dfrac{200}{201}\\ b,\dfrac{2001}{2002}=1-\dfrac{1}{2002};\dfrac{2002}{2003}=1-\dfrac{1}{2003}\\ Vì:\dfrac{1}{2002}>\dfrac{1}{2003}\Rightarrow1-\dfrac{1}{2002}< 1-\dfrac{1}{2003}\\ Vậy:\dfrac{2001}{2002}< \dfrac{2002}{2003}\)

\(c,\dfrac{2021}{2020}=1+\dfrac{1}{2020};\dfrac{2020}{2019}=1+\dfrac{1}{2019}\\ Vì:\dfrac{1}{2020}< \dfrac{1}{2019}\\ Nên:1+\dfrac{1}{2020}< 1+\dfrac{1}{2019}\\ Vậy:\dfrac{2021}{2020}< \dfrac{2020}{2019}\\ d,\dfrac{199}{198}=1+\dfrac{1}{198};\dfrac{200}{199}=1+\dfrac{1}{199}\\ Vì:\dfrac{1}{198}>\dfrac{1}{199}\\ Nên:1+\dfrac{1}{198}>1+\dfrac{1}{199}\\ Vậy:\dfrac{199}{198}>\dfrac{200}{199}\)

Bài làm:

Ta có: \(\frac{2001}{2002}< \frac{2002}{2003}\)

\(\Leftrightarrow-\frac{2001}{2002}>-\frac{2002}{2003}\)

ta có\(\dfrac{2001}{2002}< 1\)và\(\dfrac{2005}{2003}>1\)

\(\Rightarrow\dfrac{2001}{2002}< \dfrac{2005}{2003}\)

Ta có: \(\dfrac{2001}{2002}< 1\)

mà \(1< \dfrac{2005}{2003}\)

nên \(\dfrac{2001}{2002}< \dfrac{2005}{2003}\)