Ta có:

\(P=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow P=\frac{1}{1999.2000}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}\right)\)

\(\Rightarrow P=\frac{1}{1999.2000}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}\right)\)

\(\Rightarrow P=\frac{1}{1999.2000}-\left(1-\frac{1}{1999}\right)\)

\(\Rightarrow P=\frac{1}{1999.2000}-\frac{1998}{1999}\)

\(\Rightarrow P=\frac{1}{1999}-\frac{1}{2000}-\frac{1998}{1999}\)

\(\Rightarrow P=\left(\frac{1}{1999}-\frac{1998}{1999}\right)-\frac{1}{2000}\)

\(\Rightarrow P=\frac{-1997}{1999}-\frac{1}{2000}\)

\(\Rightarrow P+\frac{1997}{1999}=\frac{-1997}{1999}-\frac{1}{2000}+\frac{1}{1997}\)

\(\Rightarrow P+\frac{1997}{1999}=\frac{-1}{2000}\)

Vậy....

\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)

\(=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}+\frac{1}{1999}-\frac{1}{2000}\)

\(=\frac{1}{2}-\frac{1}{2000}=\frac{999}{2000}\)

\(P=\frac{1}{2000.1999}+\frac{1}{1999.1998}+..+\frac{1}{3.2}+\frac{1}{2.1}\)

=\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\)

=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+..+\frac{1}{1999}-\frac{1}{2000}\)

=\(1-\frac{1}{2000}\)

=\(\frac{1999}{2000}\)

$P = \dfrac1{2000 \cdot 1999} + \dfrac1{1999 \cdot 1998} + \ldots + \dfrac1{3 \cdot 2} + \dfrac1{2 \cdot 1} \\
= \dfrac1{1999} - \dfrac1{2000} + \dfrac1{1998} - \dfrac1{1999} + \ldots + \dfrac12 - \dfrac13 + \dfrac11 - \dfrac12
= - \dfrac1{2000} + \dfrac11 \\
= \dfrac{1999}{2000}$

Số hạng đầu tiên không theo quy luật hả (+) hày (-) đề thế nào làm vậy:

\(P=\frac{1}{2000.1998}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{1998.1999}\right)=\frac{1}{1999.2000}-Q\)

Tổng quát ta có \(\frac{1}{a}-\frac{1}{b}=\frac{b-a}{ab}\) với dãy trên ta luôn có b-a=1

\(Q=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1998}-\frac{1}{1999}=1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-.....-\frac{1}{1999}\)

\(Q=1-\frac{1}{1999}\Rightarrow P=\frac{1}{1999.2000}-1+\frac{1}{1999}=\frac{1-1999.2000+2000}{1999.2000}=\frac{1-1998.2000}{1999.2000}\)

\(P+\frac{1997}{1998}=\frac{1997}{1998}+\frac{1-1998.2000}{1999.2000}\) xem lại đề

Hôm kia giải thi chơi được 260, làm được bài này luôn. Hôm sau, làm lại chả biết làm.

\(P=\frac{1}{2000\times1999}-\frac{1}{1999\times1998}-\frac{1}{1998\times1997}-...-\frac{1}{3\times2}-\frac{1}{2\times1}\)

\(=\left(\frac{1}{1999}-\frac{1}{2000}\right)-\left(\frac{1}{1998}-\frac{1}{1999}\right)-\left(\frac{1}{1997}-\frac{1}{1998}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)

\(=\frac{1}{1999}-\frac{1}{2000}-\frac{1}{1998}+\frac{1}{1999}-\frac{1}{1997}+\frac{1}{1998}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(=\frac{2}{1999}-\frac{1}{2000}-1\)

3) 2x³-1=15 <=> x³=16/2=8=2³ => x=2

\(\frac{x+16}{9}=\frac{y-25}{16}=\frac{z+9}{25}=\frac{x+16+y-25+z+9}{9+16+25}=\frac{x+y+z}{50}\)

=> \(\frac{x+16}{9}=\frac{x+y+z}{50}\)=> x+y+z=\(\frac{50\left(x+16\right)}{9}\)=\(\frac{50\left(2+16\right)}{9}=\frac{50.18}{9}=50.2=100\)

Vậy x+y+z=100

Mọi người giúp tôi ik mai tôi phải thi rồi !

Mình mới học lớp 6

Nên không biết nha

Chúc các bạn học giỏi

\(P=\frac{1}{2000.1999}-\frac{1}{1999.1998}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1998.1999}+\frac{1}{1999.2000}\right)\)

\(\Rightarrow P=\frac{1}{2000.1999}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1999}-\frac{1}{2000}\right)\)

\(\Rightarrow P=\frac{1}{1999.2000}-\left(1-\frac{1}{2000}\right)\)

\(\Rightarrow P=\frac{1}{1999}-\frac{1}{2000}-\frac{1999}{2000}\)

\(\Rightarrow P=\frac{1}{1999}-1\)

\(\Rightarrow P=\frac{-1998}{1999}\)

\(\Rightarrow P+\frac{1997}{1999}=\frac{-1}{1999}\)

Vậy...

Thanks bn nha

P=(1/2000*1999)-(1/1999*1998)-...-(1/3*2)-(1/2*1)

P=(1/2000*1999)- [(1/1999*1998)+(1/1998*1997)+...+(1/2*1)]

P=(1/2000*1999)-[(1/1999)-(1/1998)+(1/1998)-(1/1997)+...+(1/2)-1]

P=(1/2000*1999)-[(1/1999)+1]

P=(1/3998000)-(2000/1999)

P=( -3999999/3998000

1023/1024 là đáp số đúng

Nguyễn Huy Thắng

Trần Việt Linh

Trương Hồng Hạnh

khỏi cần nx nhé!

"tiếng tui vang rừng núi

nhưng k ai trả lời"

có cần tui tl ?

\(\frac{1}{n\left(n+1\right)}=\frac{\left(n+1\right)-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)

\(P=\frac{1}{1999.2000}-\frac{1}{1998.1999}-...-\frac{1}{2.3}-\frac{1}{1.2}\)

\(=\frac{1}{1999}-\frac{1}{2000}-\frac{1}{1998}+\frac{1}{1999}-\frac{1}{1997}+\frac{1}{1998}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)

\(P=\frac{2}{1999}-\frac{1}{2000}-1\)

\(P+\frac{1997}{1999}=\frac{2}{1999}+\frac{1997}{1999}-\frac{1}{2000}-1=1-1-\frac{1}{2000}=-\frac{1}{2000}\)

ko biet

tiên sư mày

câu b sai đề bạn ơi

a)

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

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n}{n+1}\)

\(=\frac{1\cdot2\cdot3\cdot...\cdot n}{2\cdot3\cdot4\cdot...\cdot\left(n+1\right)}\)

\(=\frac{1}{n+1}\)

a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)

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

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n}{n+1}=\frac{1}{n+1}\)