Đề bài \(\Rightarrow3+x=1\Rightarrow x=1-3\Rightarrow x=-2\)
 

A

\(A=\dfrac{1}{3}+\dfrac{1}{6}+...+\dfrac{1}{4950}=\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{9900}=2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=2.\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=2.\dfrac{49}{100}=\dfrac{49}{50}\)

Có:  \(35x+15=890\Rightarrow35x=875\Rightarrow x=25\)

42

\(\dfrac{7}{2}\)\(:\dfrac{2}{5}\)\(=x:\dfrac{8}{7}\)

\(\dfrac{35}{4}=x:\dfrac{8}{7}\)

\(x=\dfrac{35}{4}:\dfrac{8}{7}=\dfrac{245}{32}\)

Vậy \(x=\dfrac{245}{32}\)
 

\(\dfrac{1}{k.\left(k+1\right)}\) \(=\)  \(\dfrac{1}{k}-\dfrac{1}{k+1}\)

Do

Khí nitơ