\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

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

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)

Lời giải:\(A=\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-4\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)

\(A=\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)

\(\Rightarrow A=\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right)}{2.3.4....99.100}\)

\(\Rightarrow A=\dfrac{1}{100}\)

a) Quy luật là gì ??

b) 

Đặt

 \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2020}}\\\Rightarrow2A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{2019}}\\ \Rightarrow2A-A=1-\dfrac{1}{2^{2020}}\Rightarrow A=1-\dfrac{1}{2^{2020}}\)

Suy ra , phương trình trở thành :

213 -x  =13

<=> x=200

câu thứ 2 =0 vì (63.1,-21.3,6)=0

MIK muốn hỏi câu đầu tiên

\(\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{11}-1\right)\left(\dfrac{1}{12}-1\right)...\left(\dfrac{1}{99}-1\right)\left(\dfrac{1}{100}-1\right)\)

\(=\dfrac{-9}{10}.\dfrac{-10}{11}.\dfrac{-11}{12}....\dfrac{-98}{99}.\dfrac{-99}{100}\)

\(=\dfrac{-9.-10.-11.....-98.-99}{10.11.12....99.100}\)

Vì :

Từ \(-9\) đến \(-99\) có :

\(\left(99-9\right):1+1=91\) số hạng.

Nên:

\(\)\(\dfrac{-9.-10.-11.....-98.-99}{10.11.12....99.100}=\dfrac{-\left(9.10.11....98.99\right)}{10.11.12.99.100}\)

\(=\dfrac{-9}{100}\)

\(C=\left(\dfrac{2^2-1}{2^2}\right)\left(\dfrac{1-3^2}{3^2}\right)\left(\dfrac{4^2-1}{4^2}\right)...\left(\dfrac{1-99^2}{100^2}\right)\left(\dfrac{100^2-1}{99^2}\right)=\left(\dfrac{1.3}{2^2}\right)\left(\dfrac{-2.4}{3^2}\right)\left(\dfrac{3.5}{4^2}\right)...\left(\dfrac{-98.100}{99^2}\right)\left(\dfrac{99.101}{100^2}\right)=-\dfrac{101}{200}\)

Áp dụng \(1+2+...+n=\dfrac{n\left(n+1\right)}{2}\)

\(\Rightarrow\dfrac{1}{n}\left(1+2+...+n\right)=\dfrac{n\left(n+1\right)}{2n}=\dfrac{n+1}{2}\)

Vậy:

\(A=\dfrac{1}{2}+\dfrac{3}{2}+\dfrac{4}{2}+...+\dfrac{101}{2}=\dfrac{1+2+3+...+100}{2}-1\)

\(=\dfrac{100.101}{2}-1=5049\)

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

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

\(A=\left(\dfrac{-1}{2}\right)\left(\dfrac{-2}{3}\right)\left(\dfrac{-3}{4}\right)...................\left(\dfrac{-98}{99}\right)\left(\dfrac{-99}{100}\right)\)

\(A=\dfrac{\left(-1\right)\left(-2\right)\left(-3\right).........................\left(-98\right)\left(-99\right)}{2.3.4....................98.99.100}\)

\(A=\dfrac{-1}{100}\)

Ta có

A = \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)....\left(\dfrac{1}{99}-1\right).\left(\dfrac{1}{100}-1\right)\)(99 thừa số)

A = \(\dfrac{-1}{2}.\dfrac{-2}{3}.\dfrac{-3}{4}....\dfrac{-98}{99}.\dfrac{-99}{100}\)

A = \(\dfrac{\left(-1\right).\left(-2\right).\left(-3\right)....\left(-98\right).\left(-99\right).\left(-100\right)}{2.3.4....98.99.100}\)

A = \(\dfrac{\left(-1\right).\left(-1\right).\left(-1\right)....\left(-1\right)}{1.1.1...1.1.1}\) (100 số -1, 99 số 1)

A = \(\dfrac{-1}{1.1.1.1...1.1.1}\)

A = \(\dfrac{-1}{1}\)

A = -1

Vậy A = -1