(-97)-201
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![](https://rs.olm.vn/images/avt/0.png?1311)
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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@@
![](https://rs.olm.vn/images/avt/0.png?1311)
( 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
![](https://rs.olm.vn/images/avt/0.png?1311)
`(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`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)
=>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
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
Ta có: \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\)
\(\Leftrightarrow\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)
\(\Leftrightarrow\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)
\(\Leftrightarrow\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)
mà \(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}>0\)
nên 300-x=0
hay x=300
Vậy:S={300}
(-97)-201
= (-97) + (-201)
= -298