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a) \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\)
\(\Leftrightarrow x=-2005\)
b) Sửa đề :
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)
\(\Leftrightarrow x=300\)
c) \(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\)
\(\Leftrightarrow\frac{2-x}{2002}+1=\frac{1-x}{2003}+1-\frac{x}{2004}+1\)
\(\Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\)
\(\Leftrightarrow x=2004\)
Vậy....
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\\
\)
Cộng từng hạng tử của hai vế với 1
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}=\frac{x+3+2002}{2002}+\frac{x+4+2001}{2001}\)
\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2002}=0\)
\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Vì \(\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)\ne0\)nên \(x+2005=0\Rightarrow x=-2005\)
Phương trình có nghiệm duy nhất: x=2005
(x+1)/2004+(x+2)/2003=(x+3)/2002+(x+4)/2001
(x+1)/2004+1 +(x+2)/2003 +1=(x+3)/2002+1 (x+4)/2001+1
=> x+2005/2004+(x+2005)/2003-(x+2005)/2002-(x+2005)/2002=0
(x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=>x+2005=0
=>x=-2005
[(2-x)/2001] -1 = [(1-x)/2002]-1 - [x/2003]+1
(2003-x) /2001 = (2003-x)/2002 - (2003-x)/2003
(2003-x)(1/2001-1/2002+1/2003)=0
x= 2003
mk chac chan 100% lun do
a)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\\ \Leftrightarrow\frac{201-x}{99}+\frac{99}{99}+\frac{203-x}{97}+\frac{97}{97}+\frac{205-x}{95}+\frac{95}{95}+4=4\\ \Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\) (*)
Do \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)\ne0\)
nên (*) \(\Leftrightarrow300-x=0\\ \Leftrightarrow x=300\)
b)
\(\frac{2-x}{2002}-1=\frac{1-x}{2003}-\frac{x}{2004}\\ \Leftrightarrow\frac{2-x}{2002}+\frac{2002}{2002}-1+1=\frac{1-x}{2003}+\frac{2003}{2003}-\frac{x}{2004}+\frac{2004}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}=\frac{2004-x}{2003}-\frac{2004-x}{2004}\\ \Leftrightarrow\frac{2004-x}{2002}-\frac{2004-x}{2003}+\frac{2004-x}{2004}=0\)
\(\Leftrightarrow\left(2004-x\right)\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)=0\) (*)
Do \(\left(\frac{1}{2002}-\frac{1}{2003}+\frac{1}{2004}\right)\ne0\)
nên (*) \(\Leftrightarrow2004-x=0\)
\(\Leftrightarrow x=2004\)
c) \(\left|2x-3\right|=2x-3\) (1)
ĐKXĐ: \(\\ 2x-3\ge0\)
\(\Leftrightarrow x\ge\frac{3}{2}\)
\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}2x-3=2x-3\\2x-3=-2x+3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}0x=0\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\forall x\in R\\x=\frac{3}{2}\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{\frac{3}{2}\right\}\)
\(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}\right)=\left(x-23\right)\left(\frac{1}{26}+\frac{1}{27}\right)\text{ nhận thấy:}\frac{1}{24}+\frac{1}{25}>\frac{1}{26}+\frac{1}{27}\)
\(\Rightarrow x-23=0\Leftrightarrow x=23\)
\(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\Rightarrow\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)=\left(\frac{x+3}{2002}+1\right)+\left(\frac{x+4}{2001}+1\right)\)
\(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\text{dạng giống câu a rồi nha}\)
\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\Leftrightarrow300-x=0\)
Vậy: x=300
\(c,\frac{x-a-b}{c}-1+\frac{x-b-c}{a}-1+\frac{x-a-c}{b}-1=0.\)
\(\frac{x-a-b-c}{c}+\frac{x-a-b-c}{a}+\frac{x-a-b-c}{b}=0\)
\(\left(x-a-b-c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)
=>\(\orbr{\begin{cases}a+b+c=x\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\end{cases}}\)
Vậy.......
\(a.\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\\\Leftrightarrow \left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\\\Leftrightarrow x-23=0\left(vi\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\ne0\right)\\ \Leftrightarrow x=23\)
Này tớ làm tắt có gì cậu không hiểu nói tớ nhé
\(b.\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\\ \Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1-\left(\frac{x+4}{96}+1+\frac{x+5}{95}+1\right)=0\\\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\\\Leftrightarrow \left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\\ \Leftrightarrow x+100=0\left(Vi\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\\\Leftrightarrow x=-100\)
\(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow\frac{x+2005}{2004}+\frac{x+2005}{2004}-\frac{x+2005}{2003}-\frac{x+2005}{2003}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
\(\Leftrightarrow x+2005=0\Leftrightarrow x=-2005\)
=> (x+1)/2004+1+(x+2)/2003+1=(x+3)/2002+1+(x+4)/2001+1
=> (x+2005)/2004+(x+2005)/2003=(x+2005)/2002+(x+2005)/2001
=> (x+2005)(1/2004+1/2003-1/2002-1/2001)=0
=> x+2005=0
=> x=-2005