\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\frac{1}{x+1}=\frac{1}{30}\)
\(x+1=30\)
\(x=29\)
\(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+....+\frac{1}{x\left(x+1\right)}=\frac{3}{10}\left(x\ne0;x\ne-1\right)\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{3}-\frac{1}{x+1}=\frac{3}{10}\)
\(\Leftrightarrow\frac{x+1}{3\left(x+1\right)}-\frac{3}{3\left(x+1\right)}=\frac{3}{10}\)
\(\Leftrightarrow\frac{x-2}{3\left(x+1\right)}=\frac{3}{10}\)
<=> 10(x-2)=3.3(x+1)
<=> 10x-20=9(x+1)
<=> 10x-20=9x+1
<=> 10x-20-9x-1=0
<=> x-21=0
<=> x=21 (tmđk)
Vậy x=21
