Câu hỏi của Nguyễn Gia Huy

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\)=>\(x-\dfrac{5}{13}=\dfrac{1}{3}-\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{10\cdot11}\right)\)=>\(x-\dfrac{5}{13}=\dfrac{1}{3}-\left(\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)=>\(x-\dfrac{5}{13}=\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{11}=\dfrac{1}{11}\)=>\(x=\dfrac{1}{11}+\dfrac{5}{13}=\dfrac{13}{143}+\dfrac{55}{143}=\dfrac{68}{143}\)Câu trả lời: \(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\)=>\(x-\dfrac{5}{13}=\dfrac{1}{3}-\left(\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{10\cdot11}\right)\)=>\(x-\dfrac{5}{13}=\dfrac{1}{3}-\left(\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)\)=>\(x-\dfrac{5}{13}=\dfrac{1}{3}-\dfrac{1}{3}+\dfrac{1}{11}=\dfrac{1}{11}\)=>\(x=\dfrac{1}{11}+\dfrac{5}{13}=\dfrac{13}{143}+\dfrac{55}{143}=\dfrac{68}{143}\)