Nghe có vẻ thú vị quá :> thấy hóng ghê

\(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{n.\left(n+1\right)}=\dfrac{3}{10}\)

Ta có: \(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{3}{10}\)

\(\dfrac{1}{3}-\dfrac{1}{x+1}=\dfrac{3}{10}\)

\(\dfrac{1}{x+1}=\dfrac{1}{3}-\dfrac{3}{10}\)

\(\dfrac{1}{x+1}=\dfrac{1}{30}\)

\(\Rightarrow x+1=30\)

\(x=30-1\)

\(x=29\)

Vậy ...

\(5x+1=\dfrac{6}{7}\)

\(5x=\dfrac{6}{7}-1\)

\(5x=\dfrac{-1}{7}\)

\(x=\dfrac{-1}{7}:5=\dfrac{-1}{7}\times\dfrac{1}{5}\)

\(x=\dfrac{-1}{35}\)

\(\dfrac{7}{15}.\dfrac{4}{5}+\dfrac{7}{15}.\dfrac{1}{5}-\dfrac{7}{15}\)

\(=\dfrac{7}{15}.\left(\dfrac{4}{5}+\dfrac{1}{5}-1\right)\)

\(=\dfrac{7}{15}.0\)

\(=0\)