Ta có:
N=\(\dfrac{2003+2004}{2004+2005}\)=\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)
Ta thấy:
\(\dfrac{2003}{2004+2005}\)<\(\dfrac{2003}{2004}\)(1)

\(\dfrac{2004}{2004+2005}\)<\(\dfrac{2004}{2005}\)(2)
Từ (1) và (2) --> M=\(\dfrac{2003}{2004}\)+\(\dfrac{2004}{2005}\)>\(\dfrac{2003}{2004+2005}\)+\(\dfrac{2004}{2004+2005}\)=N
Vậy  M>N

ai ra đề cho 1 lạy

HELP ME!

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

(Sửa \(cn-bm\rightarrow cn-dm\))

Ta có :

\(\left\{{}\begin{matrix}ad-bc=1\\cn-dm=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}ad=1+bc\\cn=1+dm\end{matrix}\right.\)

\(\dfrac{x}{y}=\dfrac{a}{b}.\dfrac{d}{c}=\dfrac{ad}{bc}=\dfrac{1+bc}{bc}=1+\dfrac{1}{bc}>1\left(bc>0\right)\)

\(\Rightarrow x=\dfrac{a}{b}>y=\dfrac{c}{d}\left(2\right)\)

\(\dfrac{y}{z}=\dfrac{c}{d}.\dfrac{n}{m}=\dfrac{cn}{dm}=\dfrac{1+dm}{dm}=1+\dfrac{1}{dm}>1\left(dc>0\right)\)

\(\Rightarrow y=\dfrac{c}{d}>z=\dfrac{m}{n}\left(2\right)\)

\(\left(1\right);\left(2\right)\Rightarrow x>y>z\)

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)

\(\Rightarrow x-2009=0\Rightarrow x=2009\)

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)

\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)

\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2009=0\)

\(\Leftrightarrow x=2009\)

Vậy \(x=2009\)

\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)

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

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

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

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

\(\Leftrightarrow2004-x=0\).Do \(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\ne0\)

\(\Leftrightarrow x=2004\)

Bài 1:

\(x=\dfrac{1}{2}\); \(y\) là số nguyên âm lớn nhất nên \(y=-1\). Thay x và y vào A ta được:

\(\dfrac{\dfrac{1}{2}^3-3.\dfrac{1}{2}^2+0,5.\dfrac{1}{2}-\left(-1\right)^2-4}{\dfrac{1}{2}^2+\left(-1\right)}=\dfrac{43}{6}\)

Bài 2: Tìm x

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

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

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

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

\(\Leftrightarrow x-2005=0\)

\(\Leftrightarrow x=2005\)

Vậy x=2005

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\\ \Rightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\\ \Rightarrow x=-2010\)