Câu 2:
ta có: \(\dfrac{9}{1\cdot4}+\dfrac{9}{4\cdot7}+...+\dfrac{9}{x\left(x+3\right)}=\dfrac{297}{100}\)
\(\Leftrightarrow3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{297}{100}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
=>1/x+3=1/100
=>x+3=100
hay x=97