Ta có: \(\dfrac{x}{2\cdot3}+\dfrac{x}{3\cdot4}+...+\dfrac{x}{99\cdot100}=-1\)
\(\Leftrightarrow x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=-1\)
\(\Leftrightarrow x\cdot\dfrac{49}{100}=-1\)
hay \(x=-\dfrac{100}{49}\)
$\dfrac{x}{2.3}+\dfrac{x}{3.4}+...+\dfrac{x}{99.100}=1$
`<=>x/2 - x/3 +x/3-x/4+...+x/(99)-x/(100)=1`
`<=>x/2-x/(100)=1`
`<=>(50x)/(100)-x/(100)=(100)/(100)`
`<=>50x-x=100`
`<=>49x=100`
`<=>x=(100)/(49)`
Vậy `x=(100)/(49)`
Ta có:\(\dfrac{x}{2.3}+\dfrac{x}{3.4}+.....+\dfrac{x}{99.100}=-1\)
⇔\(x\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{99.100}\right)=-1\)
⇔\(x\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{99}-\dfrac{1}{100}\right)=-1\)
⇔\(x\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=-1\)
\(\Leftrightarrow x.\dfrac{49}{100}=-1\)
\(\Leftrightarrow x=-\dfrac{100}{49}\)
Vậy phương trình có nghiệm x=\(-\dfrac{100}{49}\)