Câu trả lời:
\(\frac{1}{1X2}+\frac{1}{2X3}+\frac{1}{3X4}+\frac{1}{4X5}+\frac{1}{5X6}+\frac{1}{x}\)= \(\frac{41}{42}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{x}=\frac{41}{42}\)
\(1-\frac{1}{6}+\frac{1}{x}=\frac{41}{42}\)
\(\frac{5}{6}+\frac{1}{x}=\frac{41}{42}\)
\(\frac{1}{x}\) =\(\frac{41}{42}-\frac{5}{6}\)
\(\frac{1}{x}\) = \(\frac{1}{7}\)
VẬy x = 7
k cho mik nha