x+1/99+x+2/98+x+3/97+x+4/96=0

=> 4x+(1/99+2/98+3/97+4/96)=0

=> x=-0,025775918

giải hộ mik với min hf rất cần

Ta có : 

\(\frac{x-99-1}{99}-\frac{x-99-1}{98}-\frac{x-99-1}{97}-\frac{x-99-1}{96}-\frac{x-99-1}{95}-\frac{x-99-1}{94}=0\)

\(\Leftrightarrow\)\(\frac{x-100}{99}-\frac{x-100}{98}-\frac{x-100}{97}-\frac{x-100}{96}-\frac{x-100}{95}-\frac{x-100}{94}=0\)

\(\Leftrightarrow\)\(\left(x-100\right)\left(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)

Vì \(\frac{1}{99}-\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\)

Nên \(x-100=0\)

\(\Rightarrow\)\(x=100\)

Vậy \(x=100\)

Bài làm mang tính chất tham khảo vì em mới lớp 7 ~

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@@

( 201 - x)/99 + (203 - x)/97 + (205 - x)/95 + 3 = 0
<=> (300-x)/99+(300-x)/97+(300-x)/95=0
<=> (300-x)(1/99+1/97+1/95)=0
=> 300-x=0 vì 1/99+1/97+1/95 khác 0
=> x=300
Vậy x=300

`(201-x)/99+(203-x)/97+(205-x)/95+3=0`

`<=>(201-x)/99+1+(203-x)/97+1+(205-x)/95+1=0`

`<=>(300-x)/99+(300-x)/97+(300-x)/95=0`

`<=>(300-x)(1/99+1/97+1/95)=0`

`<=>300-x=0`

`<=>x=300`

Vậy `x=300`

Sai đề.

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

=>300-x=0

=>x=300

<=> 300-x/99+300-x/97+300-x/95=0

<=>(300-x).(1/99+1/97+1/95)=0

<=>300-x=0

<=>x=300

x=100

Ta sẽ có: 1-1+1+1-1+1-1+1=0