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

\(M=\left[\left(\frac{2}{193}-\frac{3}{386}\right).\frac{193}{17}+\frac{33}{34}\right]:\left[\left(\frac{7}{2001}+\frac{11}{4002}\right).\frac{2001}{25}+\frac{9}{2}\right] \)

     \(=\left(\frac{2}{17}-\frac{3}{34}+\frac{33}{34}\right):\left(\frac{7}{25}+\frac{11}{50}+\frac{9}{2}\right)\)

       \(=\frac{4-3+33}{34}:\frac{14+11+225}{50}=1:5=0.2\)

ta có

\(M=[(\dfrac{2}{193}-\dfrac{3}{386}).\dfrac{193}{17}+\dfrac{33}{34}]:[(\dfrac{7}{2001}+\dfrac{11}{4002}).\dfrac{2001}{25}+\dfrac{9}{2}]\)

\(\Rightarrow\)\(M=[\dfrac{1}{386}.\dfrac{193}{17}+\dfrac{33}{34}]:[\dfrac{25}{4002}.\dfrac{2001}{25}+\dfrac{9}{2}]\)

\(\Rightarrow\)\(M=[\dfrac{1}{34}+\dfrac{33}{34}]:[\dfrac{1}{2}+\dfrac{9}{2}]\)

\(\Rightarrow\)\(M=1:5\)

\(\Rightarrow M=\dfrac{1}{5}\)

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

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

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

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

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

Vậy K=2001¹⁰

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

=> x = -1999 hoặc x = - 2008