\(3,\left(9-7x\right)\left(11-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

\(4,\left(7-14x\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

\(5,\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\) 

Câu còn lại cũng tương tự

\(\dfrac{3+5x}{5}-3=\dfrac{9x-3}{4}\)

\(\Leftrightarrow\dfrac{5x-12}{5}=\dfrac{9x-3}{4}\)

\(\Leftrightarrow20x-48=45x-15\)

\(\Leftrightarrow25x=-33\)

\(\Leftrightarrow x=-\dfrac{33}{25}\)

\(A=\dfrac{1}{1\times2}+\dfrac{1}{3\times4}+...+\dfrac{1}{49\times50}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)

\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{25}\right)\)

\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)