\(B=\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)...\left(\frac{2}{2008.2009}-1\right)\)

\(B=\left(\frac{2}{2.3}-\frac{6}{2.3}\right)\left(\frac{2}{3.4}-\frac{12}{3.4}\right)...\left(\frac{2}{2008.2009}-\frac{2008.2009}{2008.2009}\right)\)

\(B=\left(-\frac{4}{2.3}\right)\left(-\frac{10}{3.4}\right)...\left(\frac{2-2008.2009}{2008.2009}\right)\)

\(B=\left(-\frac{1.4}{2.3}\right)\left(-\frac{2.5}{3.4}\right)...\left(-\frac{2007.2010}{2008.2009}\right)\)

Biểu thức B có (2008 - 2) : 1 + 1 = 2007 (thừa số)

Vì cả 2007 thừa số của biểu thức B đều mang dấu (-)

Nên biểu thức B mang dấu (-)

\(B=-\frac{1.2....2007}{2.3...2008}.\frac{4.5...2010}{3.4...2009}\)

\(B=-\frac{1}{2008}.\frac{2010}{3}\)

\(B=-\frac{1.2010}{2008.3}=-\frac{1.1005}{1004.3}=-\frac{1.335}{1004.1}\)

\(B=-\frac{335}{1004}\)

Vậy\(B=-\frac{335}{1004}\)

\(A=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot...\cdot1-\frac{2}{99\cdot100}\)

\(2A=1-\left(\frac{1}{2\cdot3}\cdot\frac{1}{3\cdot4}\cdot\frac{1}{4\cdot5}\cdot...\cdot\frac{1}{99\cdot100}\right)\)

\(2A=1-\left(\frac{1}{2}-\frac{1}{3}\cdot\frac{1}{3}-\frac{1}{4}\cdot\frac{1}{4}-\frac{1}{5}\cdot...\cdot\frac{1}{99}\cdot\frac{1}{100}\right)\)

\(2A=1-\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(2A=1-\frac{49}{100}\)

\(2A=\frac{51}{100}\)

\(A=\frac{51}{100}:2\)

\(A=\frac{51}{200}\)

\(\left(1-\frac{2}{2.3}\right)\left(1-\frac{2}{3.4}\right)\left(1-\frac{2}{4.5}\right)...\left(1-\frac{2}{99.100}\right)\)

\(=\frac{4}{2.3}.\frac{10}{3.4}.\frac{18}{4.5}...\frac{9898}{99.100}\)

\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{98.101}{99.100}\)

\(=\frac{1.2.3...98}{2.3.4...99}.\frac{4.5.6...101}{3.4.5..100}\)

\(=\frac{1}{99}.\frac{101}{3}=\frac{101}{297}\)

\(=2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)...2\left(\frac{1}{2}-\frac{2}{99.100}\right)\)
\(=2^{89}.\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)

\(=2^{98}.\left(49-\frac{49}{100}\right)\)

= \(\frac{2^{98}.4851}{100}\)

Gọi tổng trên là A
A=1/1.2.3+1/2.3.4+1/3.4.5+...1/98.99.100
Ta xét :
1/1.2 ‐ 1/2.3 = 2/1.2.3; 1/2.3 ‐ 1/3.4 = 2/2.3.4;...; 1/98.99 ‐ 1/99.100 = 2/98.99.100
tổng quát: 1/n﴾n+1﴿ ‐ 1/﴾n+1﴿﴾n+2﴿ = 2/n﴾n+1﴿﴾n+2﴿.
Do đó: 2A = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= ﴾1/1.2 ‐ 1/2.3﴿ + ﴾1/2.3 ‐ 1/3.4﴿ +...+ ﴾1/98.99 ‐ 1/99.100﴿
= 1/1.2 ‐ 1/2.3 + 1/2.3 ‐ 1/3.4 + ... + 1/98.99 ‐ 1/99.100
= 1/1.2 ‐ 1/99.100
= 1/2 ‐ 1/9900
= 4950/9900 ‐ 1/9900
= 4949/9900.
Vậy A = 4949 / 9900

Bn làm sai r . kết quả là \(\frac{101}{297}\) nhưng mik ko bt cách giải thôi

xin lỗi nha,gửi lời giải nhầm người

=\(2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)....2\left(\frac{1}{2}-\frac{1}{99.100}\right)\)

=\(2^{89}\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)

\(=2^{98}.\left(49-\frac{49}{100}\right)=\frac{2^{98}.4851}{100}\)

tham khảo câu hỏi tương tự nha bạn