a)Ta có:

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

\(=\left(\frac{1}{2.2}-1\right)\left(\frac{1}{3.3}-1\right)\left(\frac{1}{4.4}-1\right)....\left(\frac{1}{98.98}-1\right)\left(\frac{1}{99.99}-1\right)\)

\(=\left(-\frac{3}{2.2}\right).\left(-\frac{8}{3.3}\right).\left(-\frac{15}{4.4}\right)...\left(-\frac{9603}{98.98}\right).\left(-\frac{9800}{99.99}\right)\)

\(=\left[\left(-1\right).\left(-1\right).\left(-1\right)...\left(-1\right)\right].\frac{3}{2.2}.\frac{8}{3.3}.\frac{15}{4.4}...\frac{9603}{98.98}.\frac{9800}{99.99}\)

   |------------------------98 số -1--------------------|

\(=\left(-1\right)^{98}.\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)

\(=\frac{1.3}{2.3}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{95.97}{98.98}.\frac{98.100}{99.99}\)

\(=\frac{1.3.2.4.3.5...95.97.98.100}{2.2.3.3.4.4...98.98.99.99}\)

Ta sẽ rút gọn các thừa số chung ở tử và mẫu

\(=\frac{1.100}{2.99.99}\)

\(=\frac{50}{9801}\)

Vậy \(A=\frac{50}{9801}\)

cho mik hỏi b­ước 3  chỗ \(\frac{3}{2.2}\)sai o duoi lai la\(\frac{3}{2.3}\)vay

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{8\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}-\frac{8\left(x+2000\right)}{8\left(x+2000\right)\left(x+2007\right)}=\frac{7\left(x+2000\right)\left(x+2007\right)}{8\left(x+2000\right)\left(x+2007\right)}\)

\(8x+8.2007-8x+8.2000=7\left(x^2+4007x+2000.2007\right)\)

\(8.7-7\left(x^2+4007x+2000.2007\right)=0\)

\(7\left(8-x^2-4007x-2000.2007\right)=0\)

\(8-x^2-4007x-2000.2007=0\)

\(x^2+4007x+4013992=0\)

\(\left(x^2+2008x\right)+\left(1999x+4013992\right)=0\)

\(\left(x+2008\right)\left(x+1999\right)=0\)

\(\hept{\begin{cases}x=-2008\\x=-1999\end{cases}}\)

\(\frac{1}{\left(x+2000\right)\left(x+2001\right)}+\frac{1}{\left(x+2001\right)\left(x+2002\right)}+\frac{1}{\left(x+2006\right)\left(x+2007\right)}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+...+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

\(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

phần đầu mk thiếu điều kiện,bn tự bổ sung nha

a, S = 1 + 2 - 3 - 4 +5 +6 - 7 - 8 +..... +1998 -1999 -2000 +2001 
=> S = (1-3)+(2-4)+(5-7)+(6-8)+...+(1997-1999)+... + 2001 ( có 1000 hiệu = -2 ) 
=> S = -2 x 1000 + 2001 = 1 

b, S = 1 - 3 + 5 - 7 + 9 - .... - 1999 + 2001 
=> S = (1-3)+(5-7)+(9-11)+....+(1997-1999) + 2001( có 500 hiệu = -2 ) 
=> S = -2 x 500 + 2001 = 1001 

mình chỉ lmf dc 2 câu đầu thông cảm nha

\(A=2001+2001^2+...+2001^9\)

\(\Rightarrow2001A=2001^2+2001^3+...+2001^{10}\)

\(\Rightarrow2001A-A=\left(2001^2+2001^3+...+2001^{10}\right)-\left(2001+2001^2+...+2001^9\right)\)\(\Rightarrow2000A=2001^{10}-1\)

\(\Rightarrow A=\frac{2001^{10}-1}{2000}\)

\(\Rightarrow K=2000.\frac{2001^{10}-1}{2000}+1=2001^{10}-1+1=2001^{10}\)

Vậy K=2001¹⁰

Ta có:

\(M=20009.\left(2001^9+2001^8+....+2001^1\right)\)

\(2001M=20009.\left(2001^{10}+2001^9+....+2001^2\right)\)

\(\Rightarrow2001M-M=20009\left(2001^{10}+...+2001^2\right)-20009\left(2001^9+...+2001^1\right)\)

\(\Rightarrow2000M=20009\left(2001^{10}-2001^1\right)\)

\(\Rightarrow M=\dfrac{20009\left(2001^{10}-2001\right)}{2000}\)

\(\Rightarrow M=10,0045.2001\left(2001^9-1\right)\)

\(\Rightarrow M=20019,0045.\left(2001^9-1\right)\)

Chúc bạn học tốt!!!

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

=> \(\frac{1}{x+2000}-\frac{1}{x+2001}+\frac{1}{x+2001}-\frac{1}{x+2002}+....+\frac{1}{x+2006}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{1}{x+2000}-\frac{1}{x+2007}=\frac{7}{8}\)

<=> \(\frac{7}{\left(x+2000\right)\left(x+2007\right)}=\frac{7}{8}\Leftrightarrow\left(x+2000\right)\left(x+2007\right)=8\)

=> x = -1999 hoặc x = - 2008

Ta có: 2000/2001>1/2 ;  2001/2002>1/2

=>A=1/2+1/2=1=>A>1

B=2000+2001/2001+2002=4001/4003<1

A>1;B<1

=>A>B

Vậy A>B

Ta có: 2000/2001>1/2 ;  2001/2002>1/2

=>A=1/2+1/2=1=>A>1

B=2000+2001/2001+2002=4001/4003<1

A>1;B<1

=>A>B

Vậy A>B

Ta có: 2000/2001>1/2 ;  2001/2002>1/2

=>A=1/2+1/2=1=>A>1

B=2000+2001/2001+2002=4001/4003<1

A>1;B<1

=>A>B

Vậy A>B

20010

cảm ơn bạn nhưng bạn ghi rõ được cho mk ko bạn??