=> 98/97 - 97/96 + 1/96 x 1/97

=>(1/97+97/97) - (1/96+96/96) + 1/96x1/97

=>(1/97+1) - (1/96 + 1 ) +1/96x1/97

=>1/97+1-1/96-1+1/96x1/97

=>1/97-1/96+1-1 +1/96x1/97

=>1/96x1/97+1/97-1/96

đoạn sau bạn tự làm nhé

đề = 98/97 -97/96 +97/96 = 98/97

\(x=96-12=84.\)

\(x=96-12=84\)

\(x=98-78=20\)

\(x=69-27=42\)

\(x=85-20=65\)

\(x=129-85=44\)

\(x=201-129=72\)

\(x=98+129=227\)

96 - x = 12 

       x = 96 - 12 

       x = 84

78 + x = 98 

       x = 98 - 78

       x = 20

x + 27 = 69

       x = 69 - 27

       x = 42

x + 20 = 85

        x =  85 - 20 

        x =  65

x + 85 = 129

      x = 129 - 85

      x = 44

x + 129 = 201 

         x =  201 - 129

        x = 72 

x - 129 = 98 

        x = 129 - 98

        x = 31

k mình nha , Thank you ! 

12+x=96

      x=96-12

      x=84

ket qua bai tiep cung la 84

78+x=98

      x=98-78

      x=20

x+27=69

      x=69-27

      x=42

x+20=85

      x=85-20

       x=65

x+85=129

      x=129-85

      x=44

x+129=201

      x=201-129

      x=72

x-129=98

      x=98+129

      x=227

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

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

`=>(x+100)/99+(x+100)/98+(x+100)/97+(x+100)/96=0`

`=>(x+100)(1/99+1/98+1/97+1/96)=0`

`=>x+100=0` (Vì `1/99+1/98+1/97+1/96\ne0`)

`=>x=-100`

Vậy ...

`#`𝐷𝑎𝑖𝑙𝑧𝑖𝑒𝑙

\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}=-4\\ \dfrac{x+1}{99}+\dfrac{x+2}{98}+\dfrac{x+3}{97}+\dfrac{x+4}{96}+4=0\\ \left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+\left(\dfrac{x+4}{96}+1\right)=0\\ \dfrac{x+100}{99}+\dfrac{x+100}{98}+\dfrac{x+100}{97}+\dfrac{x+100}{96}=0\\ \left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+\dfrac{1}{96}\right)=0\)

mà `1/99+1/98+1/97+1/96 \ne 0`

nên `x+100=0`

`x=-100`

D

=99+(-1)=98

Chọn D

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

$\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}+4=0$

$\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0$

$\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0$

$(x+100)(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96})=0$

Vì $\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne 0$

$\Rightarrow x+100=0$

$\Rightarrow x=0-100$

$\Rightarrow x=-100$

Vậy $x=-100$

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

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

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

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

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

<=> 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}