Đặt A =\(\frac{1}{99x97}+\frac{1}{97x95}+...+\frac{1}{3x1}\)

2A =\(\frac{2}{99x97}+\frac{2}{97x95}+...+\frac{2}{3x1}\)

2A=\(\frac{1}{97}-\frac{1}{99}+\frac{1}{95}-\frac{1}{97}+...+\frac{1}{1}-\frac{1}{3}\)

2A=1-\(\frac{1}{99}\)=\(\frac{98}{99}\)

=> A=\(\frac{49}{99}\)

Đặt \(A=\frac{1}{99.97}-\frac{1}{97.95}-\frac{1}{95.93}-....-\frac{1}{5.3}-\frac{1}{3.1}\)

\(\Rightarrow A=\frac{1}{99.97}-\left(\frac{1}{1.3}+\frac{1}{3.5}+....+\frac{1}{93.95}+\frac{1}{95.97}\right)\)

\(\Rightarrow A=\frac{1}{99.97}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{95}-\frac{1}{97}\right)\)

\(\Rightarrow A=\frac{1}{99}-\frac{1}{97}-\frac{1}{2}\left(1-\frac{1}{97}\right)=\frac{1}{99}-\frac{1}{97}-\frac{1}{2}-\frac{1}{194}\)

\(\frac{-92}{93}\)

EM MỚI HỌK LỚP 6 THUI ẠK CHẮC K ĐÚNG HOẶC CÓ THỂ ĐÚNG ẠK NẾU ĐÚNG THÌ K ANH NHÓE!

CHÚC ANH HỌC TỐT

b: \(=9x^2+3x+1\)

0,2083.................

Đặt A=1/3x1 + 1/3x5 + ......+ 1/95x97 + 1/97x99

\(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)

\(2A=1-\frac{1}{99}\)

\(A=\frac{98}{99}:2\)

\(A=\frac{49}{99}\)