\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

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

\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\)

<=> x=-100

ko chép đề nhé 

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

=> \((x+100)(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95})=0\)

vì \((\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}) khác 0\)

=>\(x+100=0\)

=>x=-100

\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)

\(< =>\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)

\(< =>\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(< =>\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

\(< =>\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

\(Do:\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

\(< =>x+100=0\)

\(< =>x=-100\)

Vậy nghiệm của phương trình trên là {-100}

(=)(x+2)/98+1+(x+3)/97+1=(x+4)/96+1+(x+5)/95+1

(=)(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95

(=)(x+100)(1/98+1/97-1/96-1/95)=0

=)x+100=0

=)x=-100

\(\frac{x-1}{99}+\frac{x-2}{98}+\frac{x-3}{97}+\frac{x-4}{96}+\frac{x-5}{95}=5\)

\(\Rightarrow\left(\frac{x-1}{99}-1\right)+\left(\frac{x-2}{98}-1\right)+\left(\frac{x-3}{97}-1\right)+\left(\frac{x-4}{96}-1\right)+\left(\frac{x-5}{95}-1\right)\)\(=5-1-1-1-1-1\)

\(\Rightarrow\frac{x-100}{99}+\frac{x-100}{98}+\frac{x-100}{97}+\frac{x-100}{96}+\frac{x-100}{95}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}+\frac{1}{95}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy x=100

Chúc bạn học tốt

sửa đề đến đây thôi bạn nhé, do nếu thêm vào thì mình cũng ko biết có quy luật gì nữa :<

\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)

\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)

\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)

\(\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)

\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)

\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

=> x + 100 = 0

=> x          = -100

Vậy x = -100

Ai tra loi nhanh nhat minh se cho

4x + (1/99+2/98+3/97 + 4/96)=-4

4x=-4 - (1/99+2/98+3/97 + 4/96)

4x=

       \(\frac{2x+1}{4-x}=\frac{x-1}{5}\)

\(\Rightarrow\)\(5.\left(2x+1\right)=\left(4-x\right).\left(x-1\right)\)

\(\Rightarrow\)\(10x+5=2x-4\)

\(\Rightarrow\)\(10x-2x=-4-5\)

\(\Rightarrow\)\(8x=-9\)

\(\Rightarrow\)\(x=-\frac{9}{8}\)

(1+1+1+.............+1+1):50+2010-12xX=91

(1x50):50+2010-12xX

50:50+2010-12xX=91

1+2010-12xX=91

2011-12xX=91

12xX=2011-91

12xX=1920

x=1920:12

x=160

ngu vc

\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)

=> \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)

=> \(\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)

=> \(\left(x+99\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

=> \(x+99=0\) (Vì: \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\) )

=>\(x=-99\)

Ta có :

\(\frac{x+1}{98}+\frac{x+2}{97}=\frac{x+3}{96}+\frac{x+4}{95}\)

\(\Rightarrow\)  \(\left(\frac{x+1}{98}+1\right)+\left(\frac{x+2}{97}+1\right)=\left(\frac{x+3}{96}+1\right)+\left(\frac{x+4}{95}+1\right)\)

\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}=\frac{x+99}{96}+\frac{x+99}{95}\)

\(\Rightarrow\frac{x+99}{98}+\frac{x+99}{97}-\frac{x+99}{96}-\frac{x+99}{95}=0\)

\(\Rightarrow\left(x+99\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Vì  \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\)

\(\Rightarrow\)  \(\frac{1}{96}+\frac{1}{97}< \frac{1}{96}+\frac{1}{95}\ne0\)

Nên  \(x+99=0\)

\(\Rightarrow x=0-99\)

\(\Rightarrow x=-99\)

Vậy :        \(x=-99\)

sai đề rồi bạn ơi