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

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

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

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

Vì\(\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)\)<0 nên phương trinh đã cho tương đương:

x+2005=0 ⇔x=-2005

b) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\) 

⇔ \(\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

⇔ \(\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

⇔ \(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)>0\) nên phương trình đã cho tương đương:

300-x=0 ⇔ x=300

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}\)

\(\Leftrightarrow\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}=0\)

\(\Leftrightarrow\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\right)=0\)

\(\Leftrightarrow x-23=0\left(vì\frac{1}{24}+\frac{1}{25}-\frac{1}{26}\ne0\right)\)

\(\Leftrightarrow x=23\)

vậy................

\(\frac{201-x}{99}+\frac{203-x}{97}+\frac{205-x}{95}+3=0\)

\(\Leftrightarrow\left(\frac{201-x}{99}+1\right)+\left(\frac{203-x}{97}+1\right)+\left(\frac{205-x}{95}+1\right)=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\left(vì\frac{1}{99}+\frac{1}{97}+\frac{1}{95}>0\right)\)

\(\Leftrightarrow x=300\)

vậy..........

44

tuổi ủng hộ mk nha

xin loi minh moi hok lop 6

1. là -2005

2 làm tròn là 49,8

a) \(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)

\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}-\dfrac{x+100}{96}-\dfrac{x+100}{95}=0\)

\(\Rightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\ne0\) nên \(x+100=0\Leftrightarrow x=-100\)

b) \(\dfrac{x+1}{1998}+\dfrac{x+2}{1997}=\dfrac{x+3}{1996}+\dfrac{x+4}{1995}\)

\(\Rightarrow\dfrac{x+1}{1998}+1+\dfrac{x+2}{1997}+1=\dfrac{x+3}{1996}+1+\dfrac{x+4}{1995}+1\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}=\dfrac{x+1999}{1996}+\dfrac{x+1999}{1995}\)

\(\Rightarrow\dfrac{x+1999}{1998}+\dfrac{x+1999}{1997}-\dfrac{x+1999}{1996}-\dfrac{x+1999}{1995}=0\)

\(\Rightarrow\left(x+1999\right)\left(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\right)=0\)

Vì \(\dfrac{1}{1998}+\dfrac{1}{1997}-\dfrac{1}{1996}-\dfrac{1}{1995}\ne0\) nên \(x+1999=0\Leftrightarrow x=-1999\)

c) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Rightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Rightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Rightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\) nên \(300-x=0\Leftrightarrow x=300\)

Giải:

Ta có: \(\frac{201-x}{99}+\frac{205-x}{95}+\frac{203-x}{97}+3=0\)

\(\Leftrightarrow\frac{201-x}{99}+1+\frac{205-x}{95}+1+\frac{203-x}{97}+1=0\)

\(\Leftrightarrow\frac{300-x}{99}+\frac{300-x}{95}+\frac{300-x}{97}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\frac{1}{99}+\frac{1}{95}+\frac{1}{97}\right)=0\)

\(\Leftrightarrow300-x=0\) (Vì \(\frac{1}{99}+\frac{1}{95}+\frac{1}{97}\ne0\))

\(\Leftrightarrow x=300\)

Vậy phương trình có nghiệm là \(x=300.\)

Chúc bạn học tốt@@

Bài 1/a

Bài 1.

b) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)

\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

\(\Leftrightarrow300-x=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\ne0\))

\(\Leftrightarrow x=300\)

Vậy ....