\(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{4}{15}\)( đã sửa đề từ x+(x+1) thành x.(x+1) )
\(\Rightarrow\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4}{15}\)( Áp dụng \(\frac{1}{a.\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\))
\(\Rightarrow\frac{1}{3}-\frac{1}{x+1}=\frac{4}{15}\)
\(\Rightarrow\frac{x+1-3}{3.\left(x+1\right)}=\frac{4}{15}\)
\(\Rightarrow\frac{x-2}{3x+3}=\frac{4}{15}\)
\(\Rightarrow4.\left(3x+3\right)=15.\left(x-2\right)\)
\(\Rightarrow12x+12=15x-30\)
\(\Rightarrow12x=15x-30-12\)
\(\Rightarrow12x-15x=-30-12\)
\(\Rightarrow-3x=-42\)
\(\Rightarrow x=14\)
