Ta có :

\(A=\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)

\(A=9+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\)

\(A=\left(\frac{8}{2}+1\right)+\left(\frac{7}{3}+1\right)+...+\left(\frac{1}{9}+1\right)+1\)

\(A=\frac{10}{2}+\frac{10}{3}+...+\frac{10}{9}+\frac{10}{10}\)

\(A=10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{10\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{10}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}}=10\)

Bài 2 :

\(\frac{1}{2.3}x+\frac{1}{3.4}x+\frac{1}{4.5}x+...+\frac{1}{49.50}x=1\)

\(x\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)

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

\(x\left(\frac{1}{2}-\frac{1}{50}\right)=1\)

\(\left(\frac{25}{50}-\frac{1}{50}\right)x=1\)

\(\frac{12}{25}x=1\)

\(x=1:\frac{12}{25}\)

\(x=\frac{25}{12}\)

Su-khô-thay và Lan Xang bạn nhé

Chúc bạn học tốt

góc nhọn

các bạn làm nhanh cho mình nhé

\(\frac{1}{3}-\frac{1}{7}+\frac{3}{2}+\frac{17}{3}-\frac{27}{3}\)

\(=\left(\frac{1}{3}+\frac{17}{3}-\frac{27}{3}\right)-\left(\frac{1}{7}-\frac{3}{2}\right)\)

\(=\frac{-9}{3}+\frac{19}{14}\)

\(=\frac{-23}{14}\)