vì tử của tất cả các số là 1-1 mà 1-1=0

suy ra:=0+0+0+...+0 (100 số 0)

Suy ra:=0

vậy (1-1/1+2).(1-1/1+2+3).....(1-1/1+2+3+...+99+100)=0

M = \(\dfrac{1}{1x2}+\dfrac{1}{2x3}+\dfrac{1}{3x4}+...+\dfrac{1}{99x100}\)

M = \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

M = \(1-\dfrac{1}{100}\)

M = \(\dfrac{99}{100}\)

\(M=1\times\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{3}+\dfrac{1}{3}\times\dfrac{1}{4}+...+\dfrac{1}{99}\times\dfrac{1}{100}\)

\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

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

\(M=\dfrac{99}{100}\)

??????????

Ko bit làm 

không biết giải

2001 

____

1991

1/1*2+1/2*3+1/3*4+1/4*5+...1/99+100

Đặt \(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+....+\frac{1}{99.100}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(\Rightarrow A=\left(1+\frac{1}{2}+...+\frac{1}{100}\right)-\left(1+\frac{1}{2}+...+\frac{1}{50}\right)\)

\(\Rightarrow A=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

\(\Rightarrow A-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)

\(\Rightarrow\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)-\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=0\)

Ta có : 

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

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

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

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

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

Ủng hộ mk nha !!! ^_^

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

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

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

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

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

\(\frac{101}{297}\)\(nha\)\(bạn\)

( 1/2 - 1 ) x ( 1/3 - 1 ) x ( 1/4 - 1) x ... x ( 1/99 - 1 ) x ( 1/100 - 1 )

Từ 1/2 tới 1/100 có 99 phân số . Mà các phép tính trong các dấu ngoặc đều có kêt quả là âm

=> Kết quả của phép tính này mang giá trị âm

= - 1/2 x 2/3 x 3/4 x ... x 98/99 x 99/100

= -1/100

\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times...\times1\frac{1}{100}\)

\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times...\times\frac{101}{100}\)

\(=\frac{3\times4\times5\times...\times101}{2\times3\times4\times...\times100}\)

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