1+1=3 :)))

345,345678

Xét : \(\frac{1}{100}-\frac{1}{n^2}=\frac{n^2-100}{100n^2}=\frac{\left(n-10\right)\left(n+10\right)}{100n^2}\)

Áp dụng , đặt biểu thức cần tính là A , ta có : 

\(A=\left(\frac{1}{100}-\frac{1}{1^2}\right)\left(\frac{1}{100}-\frac{1}{2^2}\right)\left(\frac{1}{100}-\frac{1}{3^2}\right)...\left(\frac{1}{100}-\frac{1}{20^2}\right)\)

\(=\frac{\left(1-10\right)\left(1+10\right)}{100.1^2}.\frac{\left(2-10\right)\left(2+10\right)}{100.2^2}.\frac{\left(3-10\right)\left(3+10\right)}{100.3^2}...\frac{\left(10-10\right)\left(10+10\right)}{100.10^2}...\frac{\left(20-10\right)\left(20+10\right)}{100.20^2}\)

Nhận thấy trong A có một nhân tử (10-10) = 0 nên A = 0

làm thế thì hơi dài đấy Hoàng Lê Bảo Ngọc

ta nhận thấy trong biểu thức chứa thừa số \(\frac{1}{100}-\left(\frac{1}{10}\right)^2=\frac{1}{100}-\frac{1}{100}=0\)

=>biểu thức ấy =0

A=(1/100- 1^2). (1/100-(1/2)^2).....(1/100- (1/510)^2).....(1/100-(1/20)^2)

A=(1/100- 1^2). (1/100-(1/2)^2).....(1/100- 1/100).....(1/100-(1/20)^2)

A=(1/100- 1^2). (1/100-(1/2)^2).....0.....(1/100-(1/20)^2)

A=0

Mình ko biết gõ ngoặc vuông bạn thông cảm nha! Chúc bạn học tốt!!!

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)..........\left(\frac{1}{99}+1\right)\)

\(=\frac{3}{2}.\frac{4}{3}.........\frac{100}{99}\)

\(=\frac{100}{2}=50\)

\(B=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).........\left(\frac{1}{100}-1\right)\)

\(=-\frac{1}{2}.-\frac{2}{3}..........-\frac{99}{100}\)

\(=\frac{-1}{100}\)

\(A=\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)......\left(\frac{1}{99}+1\right)\)

  \(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{100}{99}\)

\(=\frac{3.4.5.....100}{2.3.4.....99}\)

 \(=\frac{100}{2}=50\)

\(\left(1+\frac{1}{100}\right).\left(1+\frac{1}{99}\right)............\left(1+\frac{1}{2}\right)\)

\(=\frac{101}{100}.\frac{100}{99}..............\frac{3}{2}\)

\(=\frac{101.100............3}{100.99...............2}\)

\(=\frac{101}{2}\)

\(=\frac{101}{100}.\frac{100}{989}.....\frac{3}{2}=\frac{101}{2}\)

Mình mới học lớp 5

mình ko trả lời được đâu nha!

a/ \(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)=\frac{3}{2}\times\frac{4}{3}\times....\times\frac{101}{100}=\frac{101}{2}\)

b/ Tự chép đề nha\(B=\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)....\left(1-\frac{1}{100}\right)\left(1+\frac{1}{100}\right)\)

\(=\frac{1}{2}\times\frac{3}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{99}{100}\times\frac{101}{100}=\frac{1}{2}\times\frac{101}{100}=\frac{101}{200}\)

Đề a) (1+1/2) (1+1/3) (1+1/4)...(1+1/100)

\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)....\left(1+\frac{1}{100}\right)\)

\(=\frac{3}{2}.\frac{4}{3}....\frac{101}{100}=\frac{3.4...101}{2.3...100}=\frac{101}{2}\)

Học tốt