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

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

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

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

k mk nha!

thanks!

nhanha!!!

Gọi A=1/99x97-1/97x95-1/95x93-...-1/5x3-1/3x1

Suy ra A=-1/1x3-1/3x5-...-1/93x95-1/95x97-1/97x99

2A=-2/1x3-2/3x5-...-1/93x95-1/95x97-1/97x99

2A=-(2/1x3+2/3x5+...+1/93x95+2/95x97+1/97x99

2A=-(1/2-1/3+1/2-1/5+...+1/93-1/95+1/95-1/97+1/97-1/99)

2A=-(1/2-1/99)

2A=-97/198

A=-97/396

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

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

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

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

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

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

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

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

A = 3/1.3 + 3/3.5 + 3/5.7 + 3/7.9 + ... + 3/97.99

A = 3/2 . ( 2/1.3 + 2/3.5 + 2/5.7 + 2/7.9 + .... + 2/97 - 2/99

A = 3/2 . ( 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99 )

A = 3/2 . ( 1 - 1/99 )

A = 3/2 . 98/99

A = 49/33

b) dãy số không có quy luật==> bạn xem lại đề

c) \(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+...+\frac{1}{50\times51}-\frac{1}{51\times52}\right)\)

\(C=\frac{1}{2}\times\left(\frac{1}{1\times2}-\frac{1}{51\times52}\right)=\frac{1}{2}\times\frac{2650}{5408}=\frac{1325}{5408}\)

\(\frac{1}{3}\times\frac{1}{6}\times\frac{1}{9}=\frac{1\times1\times1}{3\times6\times9}=\frac{1}{162}\)

\(\frac{1}{3}\times\frac{1}{6}:\frac{1}{9}=\frac{1}{3}\times\frac{1}{6}\times\frac{9}{1}=\frac{1\times1\times9}{3\times6\times1}=\frac{9}{18}=\frac{1}{2}\)

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