Question

[Lê Thị Hồng Phúc]

a) \(5\frac{8}{17}\div x+\frac{-1}{17}\div x+3\frac{1}{17}\div17\frac{1}{3}=\frac{4}{17}\)

b)\(\frac{1}{1\times4}+\frac{1}{4\times7}+\frac{1}{7\times10}+...+\frac{1}{x\times\left(x+3\right)}=\frac{6}{19}\)

Answer 1

[Thanh Tùng DZ]

a) \(5\frac{8}{17}:x+\frac{-1}{17}:x+3\frac{1}{17}:17\frac{1}{3}=\frac{4}{17}\)

\(\frac{93}{17}:x+\frac{-1}{17}:x+\frac{52}{17}:\frac{52}{3}=\frac{4}{17}\)

\(\left(\frac{93}{17}+\frac{-1}{17}\right):x+\frac{52}{17}.\frac{3}{52}=\frac{4}{17}\)

\(\frac{92}{17}:x+\frac{3}{17}=\frac{4}{17}\)

\(\frac{92}{17}:x=\frac{4}{17}-\frac{3}{17}\)

\(\frac{92}{17}:x=\frac{1}{17}\)

\(x=\frac{92}{17}:\frac{1}{17}\)

\(x=92\)

b) \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}\right)+\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{7}\right)+\frac{1}{3}.\left(\frac{1}{7}-\frac{1}{10}\right)+...+\frac{1}{3}.\left(\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{6}{19}\)

\(1-\frac{1}{x+3}=\frac{6}{19}:\frac{1}{3}\)

\(1-\frac{1}{x+3}=\frac{18}{19}\)

\(\frac{1}{x+3}=1-\frac{18}{19}\)

\(\frac{1}{x+3}=\frac{1}{19}\)

\(\Rightarrow x+3=19\)

\(\Rightarrow x=19-3\)

\(\Rightarrow x=16\)