Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+....+\frac{1}{90}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{9.10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)
\(=>\left[y-\frac{1}{2}\right]x\frac{9}{10}=\frac{1}{3}\)
\(y-\frac{1}{2}=\frac{1}{3}:\frac{9}{10}\)
\(y-\frac{1}{2}=\frac{10}{27}\)
\(=>y=\frac{10}{27}+\frac{1}{2}\)
\(y=\frac{20+27}{54}=\frac{47}{54}\)
Vậy \(y=\frac{47}{54}\)
Ủng hộ mk nha!!!
