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Đề bài tương đương:
\(\frac{201-x}{99}-1+\frac{203-x}{97}-1-\frac{205-x}{95}-1=0\)
\(\Leftrightarrow\frac{201-x}{99}-\frac{99}{99}+\frac{203-x}{97}-\frac{95}{97}-\frac{205-x}{95}-\frac{95}{95}=0\)
\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}-\frac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right).\left(\frac{1}{99}+\frac{1}{97}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow300-x=0\left(\frac{1}{99}+\frac{1}{97}-\frac{1}{95}\ne0\right)\)
\(\Leftrightarrow x=300\)
2 -x/2002 + 1 -1 = 1-x/2003 + 1 - x/2004 + 1
=> 2004 - x/ 2002 = 2004 - x/ 2003 + 2004 -x/2004
=> (2004 -x) ( 1/2002-1/2003-1/2004)
ta thấy ( 1/2002-1/2003-1/2004) # 0
=> 2004 -x = 0 => x = 2004
a, Mình nghĩ là đề sai .
b, Ta có : \(\frac{x-45}{55}+\frac{x-47}{45}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{45}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-45}{55}-\frac{55}{55}+\frac{x-47}{53}-\frac{53}{53}=\frac{x-55}{45}-\frac{45}{45}+\frac{x-53}{47}-\frac{47}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
=> \(x=100\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{100\right\}\)
c, Ta có : \(\frac{2-x}{2010}-1=\frac{1-x}{2011}-\frac{x}{2012}\)
=> \(\frac{2-x}{2010}-1=\frac{1-x}{2011}+\frac{-x}{2012}\)
=> \(\frac{2-x}{2010}+1=\frac{1-x}{2011}+1+\frac{-x}{2012}+1\)
=> \(\frac{2-x}{2010}+\frac{2010}{2010}=\frac{1-x}{2011}+\frac{2011}{2011}+\frac{-x}{2012}+\frac{2012}{2012}\)
=> \(\frac{2012-x}{2010}=\frac{2012-x}{2011}+\frac{2012-x}{2012}\)
=> \(\frac{2012-x}{2010}-\frac{2012-x}{2011}-\frac{2012-x}{2012}=0\)
=> \(\left(2012-x\right)\left(\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)=0\)
=> \(2012-x=0\)
=> \(x=2012\)
Vậy phương trình trên có tập nghiệm là \(S=\left\{2012\right\}\)
a)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)
$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$
Do đó $x-23=0\Rightarrow x=23$
b)
PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$
$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$
$\Rightarrow x+100=0\Rightarrow x=-100$
c)
PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$
Do đó $x+2005=0\Rightarrow x=-2005$
d)
PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)
\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)
\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$
Do đó $300-x=0\Rightarrow x=300$
\(\frac{201-x}{99}+\frac{203-x}{97}=\frac{205-x}{95}+3\)
\(\Leftrightarrow\frac{201-x}{99}+1+\frac{203-x}{97}+1-\frac{205-x}{95}-1=4\)
\(\Leftrightarrow\frac{200-x}{99}+\frac{200-x}{97}-\frac{200-x}{95}=4\)
\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{99}+\frac{1}{97}-\frac{1}{95}\right)=4\)
Bạn tự làm tiếp.
X = -104,695575
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